Consider $A_n$ to be a sequence of self-adjoint operators on a separable Hilbert space $H$. Moreover, assume that $A_n$ is a bounded operator for any $n$, and that $A_nx$ converges strongly to $Ax$ for $x\in S\subset H$, a dense set. I wonder whether or not $A$ is self-adjoint.
Using the Cauchy-Schwarz inequality, $$ |\langle y|A x\rangle-\langle y|A_n x\rangle |\leq\|y\| \|A x-A_nx\|, $$ and that $A_n$ converges strongly to $Ax$ for $x\in S$, we conclude that $A_nx$ also converges weakly for $x\in S$. Therefore, and since every $A_n$ is self-adjoint and bounded, we can prove that $A$ is symmetric, $$ \langle y, A x\rangle=\lim_{n\to\infty} \langle y|A_n x\rangle=\lim_{n\to\infty} \langle A_n y|x\rangle=\langle Ay, x\rangle, \quad x,y\in S. $$ In order to prove that $A$ is also self-adjoint, similarly as before we have that $$ \lim_{n\to\infty}\langle A^\dagger y-A_n y,x\rangle= 0, \quad x\in S, y\in D(A^\dagger). $$ However, can we conclude from this that the domain of $A^\dagger$, $D(A^\dagger)$ is also $S$?
Since you don't put any conditions on what $A$ does outside of $S$, you cannot expect it to be self-adjoint - self-adjointness depends very much on the domain of the operator.
To give a concrete example, take $H=L^2(0,1)$ and $A=\Delta$ on $H^2(0,1)$. This operator is not self-adjoint (not even symmetric). However, $A\vert_{C_c^2(0,1)}$ has self-adjoint extensions such as the Laplacian with Dirichlet and Neumann boundary conditions. Let $B$ be any such self-adjoint extension and $A_n=B(1-B/n)^{-1}$. The operators $A_n$ are bounded with norm $\lVert A_n\rVert\leq n$, symmetric and $A_n f\to Bf=Af$ for $f\in C_c^2(0,1)$.