Let $\tau_1$ and $\tau_2$ be two topologies on a given set $X$. We say that $\tau_1$ is coarser than $\tau_2$ if $\tau_1\subseteq \tau_2$. This means that if $U$ is open with respect to $\tau_1$ then it is open with respect to $\tau_2$. The converse need not be true.
Now, given a locally convex space $X$, $\tau_1$ being the weak topology is coarser than $\tau_2$, the original topology of $X$.
Note that the inclusion in general need not be strict. Indeed, the weak topology of a weak topology of a Banach space is the same as the weak topology of a Banach space itself.
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.
Best Answer
An affine subspace is $a+Y=\{a+y:y\in Y\}$ where $Y$ is a linear subspace of $X$.
There is exactly one topological vector space structure on a given finite dimensional real vector space.
A basis for the weak topology consists of finite intersections $V=\bigcap_{j=1}^n l_j^{-1}(U_j)$ with $l_j$ linear functions and $U_j$ are open subsets of $\Bbb R$. If nonempty, $V$ contains a subset $a+Y$ where $Y$ is the intersection of the kernels of the $l_j$, that is $Y$ is a vector subspace of finite codimension in $X$. So if $X$ is infinite-dimensional every weakly open nonempty set contains an $a+Y$ with $Y$ an infinite-dimensional vector subspace. Thus an open ball in the metric topology cannot be weakly open.