A good background reference would be Rudin's book "Functional Analysis", esp. the information on the Cayley transform and deficiency indices. The following solution is essentially the same as the previous one, just slightly more detailed.
Let $T$ be a densely defined symmetric operator on a Hilbert space $H$. Let $T$ have a unique self-adjoint extension $S$. We wish to show that the closure of $T$ is self-adjoint. Let $C$ be the closure of $T$. We wish to show that $C$ is self-adjoint. Since self-adjoint implies closed, $S$ is an extension of $C$. Since $S$ is the only self-adjoint extension of $T$, it follows that $S$ is the only self-adjoint extension of $C$. (We can now forget about $T$ and focus on $C$ and $S$.)
Let $U:=(C-i)(C+i)^{-1}$ be the Cayley transform of $C$. Let $A$ be the image of $C+i$ and let $B$ be the image of $C-i$. Then (see Rudin) $U:A\to B$ is isometric (i.e., inner product preserving) and bijective. Moreover (see Rudin), because $C$ is closed, it follows that $A$ and $B$ are both closed. Also (see Rudin), since $C$ has a unique self-adjoint extension, $U$ has a unique isometric bijective extension $H\to H$. Finally (see Rudin, vanishing of deficiency indices), to show that $C$ is self-adjoint, it suffices to show that $A=H=B$.
Let $A^\perp$ be the orthogonal complement of $A$ in $H$, and let $B^\perp$ be the orthogonal complement of $B$ in $H$. We wish to show: $A^\perp=\{0_H\}=B^\perp$.
The isometric bijective extensions $H\to H$ of $U$ are in one-to-one correspondence with isometric bijections $A^\perp\to B^\perp$. Then there is a unique isometric bijection $A^\perp\to B^\perp$. In particular, $A^\perp$ and $B^\perp$ are isometric Hilbert spaces.
For any two isometric Hilbert spaces $X$ and $Y$, unless both are zero, there are continuum many isometric bijections $X\to Y$. So, since $A^\perp$ and $B^\perp$ are isometric and only admit one isometric bijection, it follows that both are zero,
i.e., that $A^\perp=\{0_H\}=B^\perp$, as desired.