The following theorem is from Linear Operators in Hilbert Spaces, Joachim Weidmann.
Let $T$ be essentially self-adjoint on the complex Hilbert space $H$, let $S$ be symmetric with $D(T)\subseteq D(S)$ and let $\|Sf\|\leq a\|f\|+\|Tf\|$ for all $f\in D(T)$ with some $a\geq 0$. Then $T+S$ is essentially self-adjoint.
The first part of the proof:
Let $A=T+S$. We show that $R(\mu-A)^{\perp}=\{0\}$ for $\mu\in\{i,-i\}$. For this, let $(t_{n})$ be a sequence from $(0,1)$ such that $t_{n}\rightarrow 1$. By Theorem 5.28 the operator $A_{n}=T+t_{n}S$ is essentially self-adjoint for any $n\in{\bf{N}}$, and we have \begin{align*} \|(A-A_{n})f\|&=(1-t_{n})\|Sf\|\\ &\leq a\|f\|+\|Tf\|-t_{n}\|Sf\|\\ &\leq a\|f\|+\|(T+t_{n}S)f\|\\ &=a\|f\|+\|A_{n}f\|. \end{align*} Now let $h\in R(\mu-A)^{\perp}$. As $A_{n}$ is essentially self-adjoint, there exists an $f_{n}\in D(A_{n})=D(A)$ such that \begin{align*} \|(\mu-A_{n})f_{n}-h\|\leq\dfrac{1}{n},~~~~n\in{\bf{N}} \end{align*}
I fail to see why such an $f_{n}$ exists, I know that $G(\overline{A_{n}})=U(G(\overline{A_{n}})^{\perp})$ or $UG(\overline{A_{n}})=G(\overline{A_{n}})^{\perp}$, where $U(x,y)=(y,-x)$, but this does not really give the approximation to that $h$.
Actually the theorem is adopted from a paper by Wust, I have looked up the original paper, it seems that the proof approach is different, so I have no way to fix the gap that I do not understand.
If $B : \mathcal{D}(B) \subset H \rightarrow H$ is a densely-defined symmetric operator that is essentially selfadjoint, then the closure of the graph of $B$ is selfadjoint. The graph norm of $B$ is $$ \|x\|_{B}^2 = \|Bx\|^2 + \|x\|^2 = \|(B\pm iI)x\|^2,\;\; x\in\mathcal{D}(B). $$ This is because the symmetry of $B$ gives \begin{align} \|(B\pm iI)x\|^2&=\|Bx\|^2+\langle Bx,\pm ix\rangle+\langle \pm ix,Bx\rangle +\|x\|^2 \\ &=\|Bx\|^2\mp i\langle Bx,x\rangle+\pm i\langle x,Bx\rangle+\|x\|^2 \\ &=\|Bx\|^2+\|x\|^2. \end{align} The closure $\overline{B}$ of $B$ is selfadjoint. Hence, for every $y\in H$, there exists $x\in\mathcal{D}(\overline{B})$ such that $(\overline{B}+iI)x=y$. So there exists $\{ x_n \}\subset\mathcal{D}(B)$ such that $$ \lim_n Bx_n = \overline{B}x,\;\; \lim_n x_n = x. $$ Hence, \begin{align} \|y-(B+iI)x_n\|^2 & = \|(\overline{B}+iI)x-(\overline{B}+iI)x_n\|^2 \\ & = \|(\overline{B}+iI)(x-x_n)\|^2 \\ & = \|\overline{B}x-\overline{B}x_n\|^2+\|x-x_n\|^2 \\ & = \|\overline{B}x-Bx_n\|^2+\|x-x_n\|^2 \rightarrow 0. \end{align}