"Any isolated point in the spectrum of a self-adjoint operator must be an eigenvalue".
Is there an easy way to see this?
The spectral theorem tells us that any self-adjoint operator is unitarily equivalent to a multiplication operator in $L^2$ for suitable measure space. And I think unitary equivalence preserves the spectrum. So this reduces to studying isolated points in the spectrum of a multiplication operator...but does it help?
You may use the characterization of the spectrum set using spectral measure. A point on the complex plane belong to then spectrum if and only if any neighbourhood of that point has nonzero spectral measure. A point $\lambda$ on the complex plane is an eigenvalue if and only if the set $\{\lambda\}$ has nonzero spectral measure. So your claim is correct.
I will prove your statement. Most of the ideas are from Konrad's Schmudgend's excelent book Unbounded Self Adjoint Operators on Hilbert Spaces, but there he proves the result as a corollary of many Theorems. Here is a self contained shorter answer which assumes only knowledge of spectral integrals and the spectral Theorem, but which nonetheless, uses ideas from Konrad's text. Enough introduction and towards proving that:
Any isolated point of the spectrum of a self adjoint operator is an eigenvalue.
Let $\lambda_0$ be an isolated point in $\sigma(A)$. By the spectral Theorem, one has that there is a spectral measure such that $A=\int \lambda dE_A $.
Now, if $E_A(\lambda=\lambda_0)\not=0$, then one has that there is $x$ such that $E_A(\lambda=\lambda_0)x\not=0$ and by taking $M_n$ a bounding sequence ($E(\cup M_n)=I$ and $M_n\subseteq M_{n+1}$ and $\lambda|_{M_n}$ is bounded), then: $$\lim_m \int_{\mathbb{R}} \lambda \chi_M dE_AE_A(\lambda=\lambda_0) x=\lambda_0 E_A(\lambda=\lambda_0)x$$ Because the limit exists, then $E_A(\lambda=\lambda_0)x$ is in the domain of $\int \lambda dE_A$ and also: $$\int \lambda dE_AE_A(\lambda=\lambda_0)x=\lambda_0 E_A(\lambda=\lambda_0)x$$ But then $y=E_A(\lambda=\lambda_0)x$ is an eigenvector and so we have an eigenvalue.
Therefore, we need only prove that $E_A(\lambda=\lambda_0)\not=0$. To help our pursuit of this, we make precise the intuitive idea that the spectrum is related to jumps in $E_A(\lambda)$.
Lemma: Let $I(f)$ denote the integral of $f$ with respect to a spectral measure $E$.
If $\lambda \in \rho(I(f))$, then there is $\varepsilon>0$ such that $E(|f-\lambda|<\varepsilon)=0$.
If $\lambda \in \sigma(I(f))$, then $E(|f-\lambda|<\varepsilon)\not=0$ for all $\varepsilon>0$.
Proof: If $\lambda\in \rho(I(f))$, then $I(f)-\lambda=I(f-\lambda)$ is invertible. Now, if $E(f=\lambda)>0$, then $(f-\lambda)$ would have $0$ as an eigenvalue from what we have proved in the beginning of this exercise. This is a contradiction. Hence, $E(f=\lambda)=0$. It is a straightforward computation that $I(g\chi_{f=\lambda})=I(g)$ for any a.e. finite $g$ (one proves for simple functions and extends the result). Now, notice that: $$\lim_n I((f-\lambda)(f-\lambda)^{-1}\chi_{M_n})x=I(1)x=x$$ Now, it also holds that: $$I((f-\lambda)(f-\lambda)^{-1} \chi_{M_n})x=\lim_k I((f-\lambda)\chi_{M_k})I((f-\lambda)^{-1}\chi_{M_n})x$$ Hence, the limit is bounded and we have that $I((f-\lambda)^{-1}\chi_{M_n})x$ is in the domain of $I(f-\lambda)$ and: $$I((f-\lambda)(f-\lambda)^{-1} \chi_{M_n})x=I((f-\lambda))I((f-\lambda)^{-1}\chi_{M_n})x$$ Returning to the original limit: $$x=\lim_n I(f-\lambda)I((f-\lambda)^{-1}\chi_{M_n})x$$ Spectral integrals are closed and if $x\in D(I(f-\lambda)^{-1})$, then $I((f-\lambda)^{-1}\chi_{M_n})x\rightarrow I((f-\lambda)^{-1})x$, so that having a closed image implies: $$x=I(f-\lambda)I((f-\lambda)^{-1})x\quad \forall x\in \mathcal{D}(I(f-\lambda)^{-1})$$ By the same token, one writes: $$x=I((f-\lambda)^{-1})I(f-\lambda)x\quad \forall x \in \mathcal{D}(I(f-\lambda))$$ In particular from the last equation $I((f-\lambda)^{-1})\subseteq I(f-\lambda)^{-1}$. and from the first, $I((f-\lambda)^{-1})$ is injective and $I((f-\lambda)^{-1})^{-1}\subseteq I(f-\lambda)$. Now, $A\subseteq B$ where $A$ and $B$ are injective implies that $A^{-1}\subseteq B^{-1}$ because for $y\in \mathcal{R}(A)=\mathcal{D}(A^{-1})$, one has $Ax=Bx=y$ which implies $A^{-1}y=B^{-1}y=x$. Hence, $I((f-\lambda)^{-1})^{-1}\subseteq I(f-\lambda)$ teaches us that $I((f-\lambda)^{-1}\subseteq I(f-\lambda)$. All in all, we have concluded that: $$(I(f)-\lambda)^{-1}=I((f-\lambda)^{-1})$$ Now, notice that for $I((f-\lambda)^{-1})$ to be bounded, then $M=\left\{|f-\lambda|^{-1}>\|I(f-\lambda)^{-1}\|+1/n\right\}$ must have spectral measure zero as: $$\|I((f-\lambda)^{-1})\|^2\|E(M)x\|^2\geq \|I((f-\lambda)^{-1}) E(M)x\|^2\geq$$ $$\int_{M}|(f-\lambda)^{-1}|^2 d\langle E(\cdot )x,x \rangle\geq (\|I((f-\lambda)^{-1})\|+1/n)^2 \|E(M)x\|^2 $$ Thus, $E(M)x=0$ for all $x$ and we have zero spectral measure, that is, upon calling $1/\varepsilon=\|I(f-\lambda)^{-1}\|+1/n$: $$ 0=E(M )=E(|f-\lambda|\leq \varepsilon)$$ This proves the first assertion. Now for the second, if $E(|f-\lambda|<\varepsilon)=0$, then we have that $(f-\lambda)^{-1}\chi_{|f-\lambda|\geq 1/\varepsilon}$ is bounded almost everywhere, so: $$I((f-\lambda)^{-1}\chi_{|f-\lambda|\geq 1/\varepsilon})$$ Is bounded and defined everywhere. Now, notice that by boundedness and for $x$ in the domain of $I(f-\lambda)$ it holds that for a bounding sequence: $$I((f-\lambda)^{-1}\chi_{|f-\lambda|\geq 1/\varepsilon})I(f-\lambda)x=$$ $$\lim_mI((f-\lambda)^{-1}\chi_{|f-\lambda|\geq 1/\varepsilon})I((f-\lambda) \chi_{M_n})x =\lim_mI(\chi_{M_n}\chi_{|f-\lambda|\geq 1/\varepsilon} )x$$ $$=\lim_m E({M_n}\cap\{|f-\lambda|\geq 1/\varepsilon\} )x=$$ $$E((\cup M_n)\cap \{|f-\lambda|\geq \varepsilon\})x=E(\cup M_n)E(\{|f-\lambda|\geq \varepsilon)x=IIx=Ix$$
Therefore, $I(f-\lambda)$ has a bounded rightinverse. By closedness, one has that because: $$\lim_m I(f-\lambda)E(M_n)I((f-\lambda)^{-1}\chi_{|f-\lambda|\geq \varepsilon})x=x$$ $$\lim E(M_n)I((f-\lambda)^{-1}\chi_{|f-\lambda|\geq \varepsilon})x=I((f-\lambda)^{-1}\chi_{|f-\lambda|\geq \varepsilon})x$$ Then for all $x$ in our Hilbert space: $$I(f-\lambda)I((f-\lambda)^{-1}\chi_{|f-\lambda|\geq \varepsilon})x=x$$ Therefore, we have a left and right bounded inverse, so that our operator is invertible.$\square$
With this lemma at hands we are almost done. Because there is $\varepsilon$ such that $[\lambda_0-\varepsilon,\lambda_0+\varepsilon]\cap \sigma(A)=\lambda_0$, then notice that for $\varepsilon'<\varepsilon$: $$[\lambda_0-\varepsilon,\lambda_0-\varepsilon']\subseteq \rho(I(\lambda))$$ In particular, each point $t$ in this compact set has a $\delta$ neighborhood such that by the lemma $E(\{\lambda:|\lambda-t|<\delta\})=0$. Because we have a compact set, we may cover it with finitely many of these open balls of zero spectral measure, say $B_k$. Notice that by writing $A\leq B$ we mean $\langle Ax,x\rangle\leq \langle Bx,x\rangle $ for all $x$: $$0\leq E([\lambda_0-\varepsilon, \lambda_0-\varepsilon'])\leq E(\cup_{k=1}^n B_{k} ) =\sum_{k=1}^n E(B_k\setminus(\cup_{j<k} B_j))\leq \sum_{k=1}^n E(B_k)=0$$ By continuity of the measure under growing sets, we obtain: $$0=\lim_{\varepsilon'\rightarrow 0^+}E[\lambda_0-\varepsilon,\lambda_0-\varepsilon']=E[\lambda_0-\varepsilon,\lambda_0)$$ With a similar argument using Heine Borel, one has that: $$0=\lim_{\varepsilon'\rightarrow 0^+}E[\lambda_0+\varepsilon',\lambda_0+\varepsilon]= E(\lambda_0,\lambda_0+\varepsilon]$$ All in all $0\not=E([\lambda_0-\varepsilon,\lambda_0+\varepsilon])=E(\{\lambda_0\})$.