In this answer there was used the dominated convergence theorem. However, I don't see how it works here. It was said that it can be used with respect to the counting measure because the function $f$ is bounded.
Let me restate my question in more general terms:
Let $M$ be a countable subset of $\Bbb R$ and $f$ be bounded and continuous. Let $(a(k))_{k\in M}$ and $(a_n (k) )_{k\in M}$ be sequences of non-negative numbers for fixed $n$, such that $\sum_{k\in M} a_n (k) = 1, \sum_{k\in M} a(k) = 1$. Moreover, we assume that $a_n(k) \to a(k)$ for fixed $k\in M$. As I understood the answer above we then want to infer that $$\sum_{k\in M} f(k) a_n (k) = \int_M g_n (k) \text d \chi (k) \to \int_M g (k) \text d \chi (k) = \sum_{k\in M} f(k) a (k)$$ where $g_n (k) := f(k) a_n (k)$ and $g(k) := f(k)a(k)$, and $\chi $ denotes the counting measure.
However, the counting measure is not a finite measure, thus $\vert g \vert \leq \vert f \vert$ is not useful here because $f$ is not integrable, though it is bounded. The integrability here comes from the sequences $a_n$ and $ a$. But how to make use of it?
Here is a proof of weak convergence: $\sum (a(k)-a_n(k))^{+} \to 0$ by DCT because $0 \leq \left(a(k)-a_n(k)\right)^{+} \leq a_k$ and $\sum a_k =1<\infty$. But $\sum (a(k)-a_n(k))=1-1=0$ so we get $\sum (a(k)-a_n(k))^{-} \to 0$. [Because $x^{-}=x^{+}-x$]. Adding the two we get $\sum |a_n(k)-a(k)| \to 0$. It follows easily that $\sum_{k\leq j} (a_n(k)-a(k)) \to 0$ which means $P(X_n \leq j) \to P(X \leq j)$ for each $j$. So $X_n \to X$ weakly.