We show below using elementary arguments that $\ell_{\infty} / c_{0}$ has a closed subspace that is isomorphic to the nonreflexive Banach space $c_{0}(\mathbb{R})$, from which it follows that $\ell_{\infty} / c_{0}$ is not reflexive. This argument is based on Problem 2.10(c) of Topics in Banach Space Theory by Albiac and Kalton.
We first make some definitions. Let $\Gamma$ be a set.
Define $c_{00}(\Gamma )$ as the set of functions $f\colon \Gamma \to \mathbb{K}$ such that $\{\gamma \in \Gamma : f(\gamma ) \neq 0 \}$ is finite. We equip $c_{00}(\Gamma )$ with the norm $\|\cdot\|_{c_{00}(\Gamma )} \colon c_{00}(\Gamma ) \to [0, \infty )$ defined by
\begin{equation}
\|f\|_{c_{00}(\Gamma )} := \sup\{|x(\gamma )| : \gamma \in \Gamma \} .
\end{equation}
Note that $\|\cdot\|_{c_{00}(\Gamma )}$ is not a complete norm on $c_{00}(\Gamma )$ unless $\Gamma$ is finite. We only consider $c_{00}(\Gamma )$ as a normed space equipped with the norm $\|\cdot\|_{c_{00}(\Gamma )}$ in this answer.
Define $c_{0}(\Gamma )$ as the set of functions $f\colon \Gamma \to \mathbb{K}$ such that for every $\varepsilon > 0$ the set $\{\gamma\in \Gamma : |f(\gamma )| > \varepsilon \}$ is finite. We equip $c_{0}(\Gamma )$ with the norm $\|\cdot\|_{c_{0}(\Gamma )} \colon c_{0}(\Gamma ) \to [0, \infty )$ defined by
\begin{equation}
\|f\|_{c_{0}(\Gamma )} := \sup\{|f(\gamma )| : \gamma \in \Gamma \} .
\end{equation}
It is straightforward to show that is a complete norm on $c_{0}(\Gamma )$ and that $c_{00}(\Gamma )$ is dense in $(c_{0}(\Gamma ), \|\cdot\|_{c_{0}(\Gamma )})$.
Proposition. There is a closed subspace of $\ell_{\infty} / c_{0}$ that is isometrically isomorphic to $c_{0}(\mathbb{R})$.
Proof. Let $\{S_{\alpha} : \alpha \in \mathbb{R}\}$ be a collection of subsets of $\mathbb{N}$ such that $S_{\alpha}$ is infinite for each $\alpha \in \mathbb{R}$ and $S_{\alpha} \cap S_{\beta}$ is finite for any $\alpha , \beta \in \mathbb{R}$ with $\alpha \neq \beta$. See here for the existence of such a collection.
We first show that for any nonempty finite $F\subseteq \mathbb{R}$ and $\{\lambda_{\alpha} : \alpha \in F\} \subseteq \mathbb{K}$ we have
\begin{equation}
\Big\| \sum_{\alpha\in F} \lambda_{\alpha} 1_{S_{\alpha}} + c_{0} \Big\|_{\ell_{\infty} / c_{0} } = \max\{|\lambda_{\alpha}| : \alpha \in F\} . \tag{1}
\end{equation}
Fix a nonempty finite set $F\subseteq \mathbb{R}$ and scalars $\{\lambda_{\alpha} : \alpha \in F\} \subseteq \mathbb{K}$. Define $f := \sum_{\alpha\in F} \lambda_{\alpha} 1_{S_{\alpha}}$. Let $\alpha_{0}\in F$ and $\varepsilon > 0$. Take $g\in c_{0}$. For $k\in\mathbb{N}$ we consider the following conditions.
$({\rm i})$ We have $k\in S_{\alpha_{0}}$.
$({\rm ii})$ We have $k\not\in S_{\alpha} \cap S_{\beta}$ for all $\alpha , \beta \in F$ with $\alpha \neq \beta$.
$({\rm iii})$ We have $|g(k)| \leq \varepsilon$.
We remark that there are infinitely many $k\in\mathbb{N}$ such that $({\rm i})$ holds and there are only finitely many $k\in\mathbb{N}$ such that at least one of $({\rm ii})$ or $({\rm iii})$ fails, so there are (infinitely) many $k\in\mathbb{N}$ that satisfy all of $({\rm i})$, $({\rm ii})$ and $({\rm iii})$. Fix such a $k\in\mathbb{N}$ satisfying $({\rm i})$, $({\rm ii})$ and $({\rm iii})$. Then we have
\begin{equation}
|(f-g)(k)| ={} \Big| \Big( \sum_{\alpha \in F} \lambda_{\alpha} 1_{S_{\alpha}}(k) \Big) - g(k) \Big| = |\lambda_{\alpha_{0}} - g(k)| \geq |\lambda_{\alpha_{0}}| - |g(k)| \geq |\lambda_{\alpha_{0}}| - \varepsilon .
\end{equation}
Hence $\|f - g\|_{\ell_{\infty}} \geq |\lambda_{\alpha_{0}}| - \varepsilon$. Since $g\in c_{0}$ was arbitrary we obtain $\|f\|_{\ell_{\infty} / c_{0} } \geq |\lambda_{\alpha_{0}}| - \varepsilon$. It follows that $\|f + c_{0}\|_{\ell_{\infty} / c_{0} } \geq \max\{|\lambda_{\alpha}| : \alpha \in F\}$, that is
\begin{equation}
\Big\| \sum_{\alpha\in F} \lambda_{\alpha} 1_{S_{\alpha}} + c_{0} \Big\|_{\ell_{\infty} / c_{0} } \geq \max\{|\lambda_{\alpha}| : \alpha \in F\} . \tag{2}
\end{equation}
Conversely, again taking $f := \sum_{\alpha\in F} \lambda_{\alpha} 1_{S_{\alpha}}$, since there are finitely many $k\in\mathbb{N}$ such that $k\in S_{\alpha} \cap S_{\beta}$ for some $\alpha , \beta \in F$ with $\alpha \neq \beta$ we can write $f = f_{1} + f_{2}$ where $f_{1}\in \ell_{\infty}$ has range $\{\lambda_{\alpha} : \alpha \in F\}$ and $f_{2}\in c_{00}$. Then
\begin{equation}
\|f - f_{2}\|_{\ell_{\infty}} = \|f_{1}\|_{\ell_{\infty}} = \max\{|\lambda_{\alpha}| : \alpha \in F\} .
\end{equation}
It follows that $\|f + c_{0}\|_{\ell_{\infty} / c_{0}} \leq \max\{|\lambda_{\alpha}| : \alpha \in F\}$, that is
\begin{equation}
\Big\| \sum_{\alpha\in F} \lambda_{\alpha} 1_{S_{\alpha}} + c_{0} \Big\|_{\ell_{\infty} / c_{0} } \leq \max\{|\lambda_{\alpha}| : \alpha \in F\} . \tag{3}
\end{equation}
By combining $(2)$ and $(3)$ we obtain $(1)$.
Define $\Lambda_{0} \colon c_{00}(\mathbb{R}) \to \ell_{\infty} / c_{0}$ by
\begin{equation}
\Lambda_{0}((\lambda_{\alpha})_{\alpha\in\mathbb{R}} ) = \sum_{\alpha\in\mathbb{R}} \lambda_{\alpha} 1_{S_{\alpha}} + c_{0} .
\end{equation}
By $(1)$ we see that $\Lambda_{0} \colon (c_{00}(\mathbb{R}), \|\cdot\|_{c_{00}(\mathbb{R})}) \to \ell_{\infty} / c_{0}$ is a linear isometry. Then as $c_{00}(\mathbb{R})$ is dense in $(c_{0}(\mathbb{R}), \|\cdot\|_{c_{0}(\mathbb{R})})$, it follows by continuity that there is a continuous linear isometry $\Lambda \colon c_{0}(\mathbb{R}) \to \ell_{\infty} / c_{0}$ such that $\Lambda\vert_{c_{00}(\mathbb{R})} = \Lambda_{0}$. It follows that $c_{0}(\mathbb{R})$ is isometrically isomorphic to the closed subspace ${\rm ran} \, \Lambda$ of $\ell_{\infty} / c_{0}$.
We now use the preceding result to deduce that $\ell_{\infty} / c_{0}$ is not reflexive.
Corollary. The space $\ell_{\infty} / c_{0}$ is not reflexive.
Proof. Define $M := \{f\in c_{0}(\mathbb{R}) : f(\alpha ) = 0 \text{ for all } \alpha \in \mathbb{R}\setminus \mathbb{N} \}$. Then $M$ is a closed subspace of $c_{0}(\mathbb{R})$ and the map $f\mapsto f\vert_{\mathbb{N}}$ is an isometric isomorphism from $M$ onto $c_{0}$. Hence $c_{0}(\mathbb{R})$ has a closed subspace that is isomorphic to the nonreflexive Banach space $c_{0}$. Consequently, $c_{0}(\mathbb{R})$ is not reflexive. Since we have previously shown that $\ell_{\infty} / c_{0}$ has a closed subspace isomorphic to $c_{0}(\mathbb{R})$, we conclude that $\ell_{\infty} / c_{0}$ is not reflexive.
Here are some remarks.
The same argument can be used to show that $\ell_{\infty} / c_{0}$ fails to have properties that are passed down through closed subspaces that $c_{0}(\mathbb{R})$ fails to have. For example, since $c_{0}(\mathbb{R})$ is not separable and not weak sequentially complete, we see that $\ell_{\infty} / c_{0}$ is not separable and not weak sequentially complete.
Using a similar argument, it can be shown that $\ell_{\infty} / c_{0}$ has a closed subspace isomorphic to $\ell_{\infty}$. Let $\{S_{n} : n\in\mathbb{N}\}$ be an infinite collection of pairwise disjoint infinite subsets of $\mathbb{N}$. Define $\Phi \colon \ell_{\infty} \to \ell_{\infty}$ by
\begin{equation}
(\Phi f)(j) := \begin{cases} f(k) & \text{if } j\in S_{k} \text{ for some } k\in \mathbb{N} , \\
0 & \text{otherwise}.
\end{cases}
\end{equation}
Let $\pi \colon \ell_{\infty} \to \ell_{\infty} / c_{0}$ be the natural quotient map and define $\Psi \colon \ell_{\infty} \to \ell_{\infty} / c_{0}$ by $\Psi := \pi \circ \Phi$. Using similar methods to the proof of the proposition in this answer, it can be shown that $\Psi$ is a linear isometry. Hence $\Psi (\ell_{\infty})$ is a closed subspace of $\ell_{\infty} / c_{0}$ that is isometrically isomorphic to $\ell_{\infty}$. Then since $\ell_{\infty}$ is not reflexive it follows that $\ell_{\infty} / c_{0}$ is also not reflexive.
