Does weak $ L^2$ convergence $+$ (uniform) $H^1$ boundedness imply weak $H^1$ convergence on bounded domains? I'm really surprised I can't seem to find this statement anywhere, even though it seems to follow from compactness theorems, I'm therefore wondering whether I made a mistake in my proof below. 
Claim: Let $\Omega$ be a sufficiently nice bounded domain. Let $u_n \in H^1(\Omega)$ with $u_n \rightharpoonup u$ in $L^2$, where $\rightharpoonup$ denotes weak convergence. Assume further that $\| \nabla u_n\|_{L^2} \leq C$ for some $C > 0$ independent of $n$. Then $u_n \rightharpoonup u$ in $H^1$.
Proof: Since $u_n$ is bounded in $H^1$, it must have a $H^1$-weakly convergent subsequence to some $\tilde u \in H^1$  by the sequential Banach-Alaoglu theorem. Since this subsequence is also $L^2$-weakly convergent and weak limits are unique, it is true that $u = \tilde u$ and in particular $u \in H^1$. Now assume that the sequence $u_n$ does not converge weakly to $u$. Then there must be some subsequence that is completely outside of a neighborhood $U$ of $u$. By the same compactness argument, this subsequence again has a $H^1$-weakly convergent subsequence which must converge to $u$, and hence be in $U$ infinitely often, a contradiction.

The result is true and your argument is correct. A slightly better phrasing might be to show that every subsequence of $u_n$ has a further subsequence that converges weakly to $u$ in $H^1$. It is then a standard topological result that this implies that $u_n  \rightharpoonup u$. In fact, the last part of your argument would essentially prove the general topological result.
Perhaps the reason that you cannot find this exact result anywhere is that it is a not too difficult instance of the well known criterion that says that a relatively compact sequence with a unique limit point must converge to that limit point. 
