Let $H$ be a $\mathbb C$-Hilbert space and $T$ be a densely-defined symmetric linear operator on $H$.
Can we show that $T^{\ast\ast}\subseteq T^\ast$ without using the fact that $T$ is closable and $\overline T$ is symmetric as well?
I know how we can easily prove the claim using the mentioned fact, since $$T^{\ast\ast}=\overline T\subseteq\overline T^\ast=T^\ast\tag1.$$ However, I've found the claim in several books at stages where the authors haven't defined the notion of the closure yet and since all of them mention the claim without any further remark I've got the strange feeling that the desired relation is way more obvious from the symmetry than I realize up to now.
Since $T$ is symmetric you can check explicitly that $D(T)\subseteq D(T^*)$ and $T^*\lvert_{D(T)}=T$.
You can then prove $T^{**}\subseteq T^*$ from the definition:
Let $x\in D(T^{**})$, then there is a $C$ so that $|(x,T^{*}y)| ≤ C\,\|y\|$ for all $y\in D(T^*)$. In particular it holds for all $y\in D(T)$, where $T^*$ and $T$ agree, so the inequality becomes: $$|(x,Ty)|≤ C\|y\| \ \ \forall y\in D(T)$$ giving that $x\in D(T^{*})$. Now $T^{**}x$ is determined by the extension of functional $y\mapsto (T^{**}x,y) = (x,Ty)$. But that is the same functional determining $T^*x$ (one initial domain is bigger than the other, but both are dense so the extensions to all of $H$ agree), hence you also have $T^{**}x=T^*x$ whenever $x\in D(T^{**})$.