In general a topological vector space doesn't have that property.
An example in $\ell^2(\mathbb{N})$ in its weak topology:
For $m \in \mathbb{N}\setminus \{0\}$, let
$$A_m = \bigl\{ x \in \ell^2(\mathbb{N}) : \lVert x\rVert_2 = m, \bigl(n \leqslant \tfrac{m(m-1)}{2} \lor n > \tfrac{m(m+1)}{2}\bigr) \implies x_n = 0\bigr\}$$
and define
$$A = \bigcup_{m = 1}^{\infty} A_m\,.$$
Then the intersection of $A$ with every closed and bounded set is compact (even in the strong topology), thus all accumulation points of bounded nets in $A$ lie in $A$, in particular if a bounded net in $A$ converges to $x_0$, then $x_0 \in A$.
But $A$ is not weakly closed, we have $0 \in \operatorname{cl}_w(A) \setminus A$. For every weak neighbourhood of $0$ contains one of the form
$$V(\varepsilon;\xi_1, \dotsc, \xi_k) = \{ x \in \ell^2(\mathbb{N}) : \lvert\langle x, \xi_j\rangle\rvert < \varepsilon \text{ for } 1 \leqslant j \leqslant k\} $$
where $k \in \mathbb{N}$, $\xi_1,\dotsc,\xi_k \in \ell^2(\mathbb{N})$, and $\varepsilon > 0$. And $V(\varepsilon;\xi_1,\dotsc,\xi_k) \cap A_m \neq \varnothing$ for all $m > k$.
This construction can be imitated in every infinite-dimensional normed space and yields a set that isn't weakly closed but whose intersection with every weakly closed bounded set is weakly closed.
The property holds (as you know) in every metrisable topological vector space.
For convex $A$ we have the equivalence
$$A\text{ closed} \iff (A\cap B)\text{ closed for all closed and bounded } B$$
if $X$ carries the weak topology of an originally locally convex and metrisable space. In particular in the weak topology of a Banach space, closed convex sets can be characterised by the convergence of bounded nets.
