Consider the sequence of real-valued random variables $\{X_n\}_n$ and suppose it converges in distribution to $V$. Does this imply that
(1) every subsequence of $\{X_n\}_n$, $\{X_{n_k}\}_k$ converges in distribution to $V$? Why?
(2) every subsequence of $\{X_n\}_n$, $\{X_{n_k}\}_k$, has a subsequence, $\{X_{n_{k_j}}\}_j$ that converges in distribution to $V$? Why?
I know by Prohorov's Theorem that $X_n \rightarrow_d V$ implies that $\{X_n\}$ is uniformly bounded in probability which implies that there exists a subsequence of $X_n$, $\{X_{n_k}\}_k$, converging to $V_k$. Moreover, since also $\{X_{n_k}\}_k$ is uniformly bounded in probability, we can apply again the Prohorov's Theorem and say that every $\{X_{n_k}\}_k$ has a subsequence $\{X_{n_{k_j}}\}_j$ that converges in distribution to $V_{k_j}$.
I don't know how to go further to show (1) or (2)
This does not require the Prokhorov's theorem: just use the fact that $Y_n\to Y$ in distribution if and only if $ \mathbb P\{Y_n \leqslant t \} \to\mathbb P\{Y\leqslant t\}$ for each $t$ such that the map $s\mapsto \mathbb P\{Y\leqslant s\}$ is continuous.
In order to check (1), you have to prove that $\mathbb P\{X_{n_k}\leqslant t\}\to \mathbb P\{X\leqslant t\}$ for each $t$ such that the map $s\mapsto \mathbb P\{X\leqslant s\}$ is continuous. Fix such a point and define $c_n:=\mathbb P\{X_n\leqslant t\}$. The claim follows from the fact that if $(c_n)$ converges to some $c$, then so does a subsequence $(X_{n_k})$ .
For (2), just take the original subsequence (no need to extract a further subsequence, by (1)).