Given $(E,\| \cdot \|_E),(F,\|\cdot \|_F)$ reflexive normed vector spaces. I have to prove that also $(E\times F,\|\cdot \|_{E\times F})$ is reflexive where $\|\cdot \|_{E\times F}$ is the product norm.
What I know is that $(E\times F)'$ is algebrically and topologically isomorphic to $E'\times F'$.
Let $J:(E'\times F')\to (E\times F)'$ be defined by $(J(e,f))(x,y)=e(x)+f(y)$. We know that $$J(E'\times F')=(E\times F)'.\tag{$*$}$$
Take a bounded sequence $(x_n,y_n)$ in $E\times F$. Then $(x_n)$ is bounded in $E$ and $(y_n)$ is bounded in $F$. Since $E$ and $F$ are reflexive, there are subsequences $(x_{n_k})$, $(y_{n_k})$ and vectors $x\in E$, $y\in F$ such that $$x_{n_k}\rightharpoonup x\text{ in }E,\quad y_{n_k}\rightharpoonup y\text{ in }F$$ and thus $$(J(e,f))(x_{n_k},y_{n_k})=e(x_{n_k})+f(y_{n_k})\to e(x)+f(y)=(J(e,f))(x,y),\quad\forall\;(e,f)\in E'\times F'.$$ It follows from $(*)$ that $$g(x_{n_k},y_{n_k})\to g(x,y),\quad \forall\; g\in (E\times F)'$$ and thus
$$(x_{n_k},y_{n_k})\rightharpoonup (x,y)\text{ in }E\times F.$$
This shows that every bounded sequence in $E\times F$ admits a weakly convergent subsequence. So, by the Eberlein-Šmulian theorem, $E\times F$ is reflexive.
Here are two more proofs.
We define $j_{E} : E^*\rightarrow (E\times F)^{*}$ by $(j_{E}(f))(x, y) = f(x)$ (and similarly define $j_{F}$). It is clear that $j_{E}$ is continuous since $$|j_{E}(f)(x, y)| = |f(x)|\leq \|f\|\|x\|\leq \|f\|(\|x\|+\|y\|) = \|f\|\|(x, y)\|.$$ Moreover, it is also clear that for any $h\in (E\times F)^{*}$ we can write $h = j_{E}(f)+j_{F}(g)$ for some (unique) $(f, g)\in E^*\times F^*$. (Here, $f(x) = h(x, 0)$ and $g(y) = h(0, y)$.) If $\Phi\in (E\times F)^{**}$ then $\Phi(h) = \Phi(j_{E}(f))+\Phi(j_{F}(g))$. Define $\varphi : E^{*}\rightarrow\mathbb{R}$ by $\varphi(f) = \Phi(j_{E}(f))$. Since $j_{E}$ is continuous, $\varphi\in E^{**}$ and using the reflexivity of $E$ we get $\varphi = \varphi_{x}$ for some $x\in E$ (where $x\mapsto \varphi_{x}$ is the natural map from $E$ to $E^{**}$). We can similarly define a map $\psi\in F^*$ and use the fact that $\psi = \psi_{y}$ for some $y\in F$. Thus, $$\Phi(h) = \varphi(f)+\psi(g) = \varphi_{x}(f)+\psi_{y}(g) = f(x)+g(y) = h(x, y) = \Phi_{(x, y)}(h).$$ This shows that $E\times F$ is reflexive.
If $E$ and $F$ are Banach spaces, then we can use a theorem by Kakutani to give another proof. The theorem states that a Banach space $E$ is reflexive if and only if the unit ball $B_{E} = \{x\in E: \|x\|\leq 1\}$ is compact in the weak topology $\sigma(E, E^*)$. Using the (equivalent) norm $\|(x, y)\| = \max\{\|x\|, \|y\|\}$ on $E\times F$ we get $B_{E\times F} = B_{E}\times B_{F}$. Once one has shown that the weak topology $\sigma(E\times F, (E\times F)^*)$ is the same as the product topology $\sigma(E, E^*)\times \sigma(F,, F^*)$ (which is also the reason why the Eberlein-Šmulian argument in the other answer works), the result then follows immediately.