Take $X = \{0,1\}^{[0,1]}$, thought of as the set of all functions from the unit interval $[0,1]$ to the two-point set $\{0,1\}$. Equip $X$ with the product topology; $X$ is compact by Tychonoff's theorem. In the product topology, a sequence $f_n$ converges iff it converges pointwise, i.e. if the $\{0,1\}$-valued sequence $f_n(x)$ converges (i.e. is eventually constant) for every $x \in [0,1]$.
Define $f_n : [0,1] \to \{0,1\}$ to be the function such that $f_n(x)$ is the $n$th bit in the binary expansion of $x$. (If $x$ has more than one binary expansion, take $f_n(x) = 0$ for all $n$; it doesn't really matter here.) Then given any subsequence $f_{n_m}$, you should be able to produce an $x \in [0,1]$ such that $f_{n_m}(x)$ does not converge. Hence $\{f_n\}$ has no convergent subsequence.
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.
Best Answer
See Theorem $15.3$ in this excellent PDF, Translating Between Nets and Filters, by Saitulaa Naranong; it’s well worth reading the whole thing.