Suppose $T$ is symmetric, and has a unique selfadjoint extension. (I'll assume that $T$ is densely-defined, though you can probably prove that it must be the case.)
Let $\overline{T}$ be the closure of $T$, which exists because $T$ is symmetric. $\overline{T}$ must also be symmetric, and because $\overline{T}^*=T^*$, then
$$
\mathcal{D}(T^*)=\mathcal{D}(\overline{T})\oplus\mathcal{N}(T^*-iI)\oplus\mathcal{N}(T^*+iI),
$$
where the decomposition is orthogonal with respect to the graph inner product
$$
(x,y)_{T^*} = (x,y)_H+(T^*x,T^*y)_{H}.
$$
If $S$ is a selfadjoint extension of $T$, then $S$ is also a selfadjoint extension of $\overline{T}$ because the graph of $T$ is contained in the graph of $S$ and the graph of $S$ is closed. $\overline{T}$ has a selfadjoint extension iff $\mathcal{N}(T^*-iI)$ is unitarily equivalent to $\mathcal{N}(T^*+iI)$, and the possible selfadjoint extensions are in one-to-one correspondence with the unitary maps $U: \mathcal{N}(T^*-iI)\rightarrow\mathcal{N}(T^*+iI)$. However, because $\overline{T}$ has only one selfadjoint extension, then $\mathcal{N}(T^*-iI)=\{0\}=\mathcal{N}(T^*+iI)$ follows. Hence, $\overline{T}=T^*$, which also gives $\overline{T}=\overline{T}^*$. So $\overline{T}$ must be selfadjoint.
