The couterpart of result (1) can fail if the sequence is replaced by the net. Our counterexample is based on Nate Eldredge’s counterexample. Direct a set $I=I’\times\Bbb N$ by the preorder $\preceq’$ defined by
$$(U’,n’) \preceq’ (V’, m’) \mbox{ iff } U’ \preceq V’ \mbox{ and } m’\ge n’.$$
For each $U\in\mathcal U$ pick $x_U\in X$ such that $\|x_U\|=1$ and $\langle f_U, xU\rangle\ne 0$. Define nets indexed by $I’$ putting $x^*_{(U,n,n’)}=f_{U,n}=nf_U$ and $x_{(U,n,n’)}=\frac 1{n’}x_U$ for for each $(U,n,n’)\in I$. Clearly, the net $\{ x_{(U,n,n’)}: (U,n,n’)\in I’\}$ converges to the zero. Since the net $\{f_{U,n}:(U,n)\in I\}$ converges to the zero, the net $\{ x^*_{(U,n,n’)}: (U,n,n’)\in I’\}$ converges to the zero too. On the other hand, for each $(U,n,n’)\in I’$ and each natural $m$ we have $(U,n,n’)\preceq’ (U,m,n’)$ and $\langle x^*_{(U,m,n’)}, x_{(U,m,n’)}\rangle=\langle mf_U, \frac 1{n’}x_U \rangle=
\frac {m}{n’} \langle f_U, x_U \rangle$, which has an absolute value bigger than $1$ for a sufficiently big $m$.
The couterpart of result (1) holds when the directed set $(I,\le)$ of the net has countable cofinalty, that is there exists a countable set $D$ of $I$ such that for each $n\in I$ there exists $d\in D$ with $d\ge n$. Indeed, suppose to the contrary that $\langle x^*_n, x_n\rangle\not\rightarrow \langle x^*,x\rangle$.
Then there exists $\varepsilon>0$ such that for each $n\in I$ there exists $n’\ge n$ such that
$|\langle x^*_n, x_n\rangle - \langle x^*,x\rangle|\ge\varepsilon$.
Let $\{d(k):k\in\Bbb N\}$ be any enumeration of the set $D$.
Then by indution we can build a sequence $\{n(k):k\in\Bbb N\}$ of elements of $I$ such that for each $k$ we $n(k)\ge d(k)$ and $|\langle x^*_{n(k)}, x_{n(k)}\rangle - \langle x^*,x\rangle|\ge\varepsilon$. But a sequence $\{x_{n(k)}\}$ converges to $x$ and a sequence $\{x^*_{n(k)}\}$ converges to $x^*$, a contradiction with result (1).
Best Answer
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.