I think what you are saying is true. Never thought about it since i've always pre-assumed that the weakly-operator limit $A$ of the $A_n's$ was always in $A\in \mathfrak L(X,Y)$. Am writing the argument just to convience ourselfs. Indeed, we only need to assume that $Y$ has a norm, not necessarily a complete one.
So, lets suppose that $A_n\overset{\text{wo}}{\to}A$ in the weak operator topology where $A:X\to Y$ is a linear operator, not necessarily bounded. Convergence in the weak operator topology is described by $h(A_n x)\to h(A x)$ for every $x\in X$ and $h\in Y^*$. This implies that the set $\{A_n x: n\in \mathbb{N}\}$ is weakly bounded in $Y$, hence it is also bounded in $Y$. By the Banach-Steinhaus it follows that $\sup_{n}||A_n||=M<\infty$. Now, for $x\in X$ with $||x||=1$ we have
$$||Ax||=\max_{h\in Y^*,\, ||h||=1}|h(Ax)|$$
So, there is some $||h||=1$ in $Y^*$ such that $||Ax||=|h(Ax)|$. Using the weak convergence for $A_nx$ we end up with
\begin{align}
||Ax||&=|h(Ax)|\\
&=\lim_{n\to \infty}|h(A_nx)|\\
&\leq \underbrace{||h||}_{=1}\liminf_{n\to \infty}||A_n||\cdot \underbrace{||x||}_{=1}
\end{align}
Hence, $||Ax||\leq M$ for every $||x||=1$ and therefore, $||A||\leq M<\infty$.
Edit: (Responding to the comment)
The existence of such $A$ is trickier. To ensure such existence we need another assumption for $Y$, since there is a counter example in here where $X=Y=c_0$. The only natural that i could think while i was trying to prove it is that $Y$ has to be reflexive (from not being a Banach space we went straight out to reflexivity :P). In the case where $X=Y=H$ is a Hilbert space things were slightly more easier since we can identify $H^*$ with $H$ and dont need to mess with the second duals.
The argument in the case where $Y$ is reflexive is the following:
Suppose that $\lim_{n}\langle A_n x, h \rangle$ exists for every $x\in X$ and $h\in Y^*$. For fixed $x\in X$ let $f_x:Y^*\to \mathbb{R}$ defined by
$$\langle h, f_x\rangle =\lim_{n\to \infty}\langle A_n x, h\rangle$$
Its easy to check that $f_x$ is a linear functional and by the previous discussion it is also bounded. Meaning, $f_x \in Y^{**}$. By reflexivity, there is some $y_x\in Y$ such that $\langle h, f_x\rangle =\langle y_x, h\rangle$ for all $h\in Y^*$. Now, let $x\overset{A}{\longmapsto} y_x$. Now, its easy to check that $A:X\to Y$ is a linear operator. By the previous discussion it is also bounded.
$$\sup_{||x|| \leq 1} ||(T_\alpha-T)x || \to 0 \implies ||(T_\alpha-T)x || \to 0\, , \, \forall x \in H \, , \, ||x|| \leq 1.\\$$
Consider $e_k$ an orthonormal basis of $H$ and define $T_n(e_k)=e_k$ for $k\leq n$ and $T_n(e_k)=2e_k$ for $k >n $.
Then $$||(T_n-T)x || =||\sum_{k=n+1}^{\infty}\langle x,e_k\rangle e_k||\to 0\, , \, \forall x \in H \, , \, ||x|| \leq 1 $$
But
$$\sup_{||x|| \leq 1} ||(T_n-T)x ||\geq ||(T_n-T)e_{n+1} ||=||e_{n+1}||=1\\$$
Best Answer
Finite dimension is alwyas a characterization: Fix $x\in X$ with norm $1$. Then $X^*\to B(X)$, $\varphi \mapsto (y\mapsto \varphi(y)x)$ identifies the dual $X^*$ with a subspace of $B(X)$ such that the operator norm of $B(X)$ induces the dual norm on $X^*$ and the SOT induces the weak$^*$-topology on $X^*$. Moreover, the weak$^*$ topology coincides with the dual norm topology if and only if $X$ is finite dimensional. (There is a weak$^*$-seminorm $p(\varphi)=\max\{|\varphi(x_1)|,\ldots,|\varphi(x_n)|\}$ majorizing the dual norm and by Hahn-Banach this implies that the linear span of $\{x_1,\ldots,x_n\}$ is $X$).