Using Prohorov's theorem to prove uniform tightness I am currently studying van der Vaart's Asymptotic statistics. I have a question on the following example in Section 2.2:

"Prove that $o_{P}(1)O_{P}(1)=o_{P}(1)$"

Let me give some definitions first, just in case:

$X_{n}=o_{P}(1)$ denotes that the sequence of random vectors $X_{n}$
converges in probability to $0$, in other words
$X_{n}\overset{P}{\to}0$, or $P(\lVert X_{n}\rVert>\varepsilon)\to0$
for all $\varepsilon>0$.
$Y_{n}=O_{P}(1)$ denotes that the sequence of random vectors $Y_{n}$
is bounded in probability, which is the same as saying that $Y_{n}$ is
uniformly tight (according to van der Vaart), namely for every
$\varepsilon>0$ exists constant $M>0$ such that
$\underset{n}{\text{sup}}\ P(\lVert Y_{n}\rVert>M)<\varepsilon$.
In addition, by $X_{n}\overset{\mathcal{D}}{\to}X$ I denote the weak convergence, or convergence in distribution of the sequence of random vectors $X_{n}$ to some random vector $X$, namely that $P(X_{n}\le x)\to P(X\le x)$.

Let me also write Prohorov's theorem (Theorem 2.4 in van der Vaart's Asymptotic statistics), as I will be referring to it:

Prohorov's theorem Let $X_{n}$ be random vectors in $\mathbb{R}^{k}$.
(i) If $X_{n}\overset{\mathcal{D}}{\to}X$ for some $X$, then
$\{X_{n}\}$ is uniformly tight;
(ii) If $X_{n}$ is uniformly tight,
then there exists a subsequence with
$X_{n_{j}}\overset{\mathcal{D}}{\to}X$ as $j\to\infty$, for some $X$.

I will now illustrate the proof of van der Vaart for $o_{P}(1)O_{P}(1)=o_{P}(1)$.
Suppose $X_{n}=o_{P}(1)$ and $Y_{n}=O_{P}(1)$. Hence, $X_{n}\overset{P}{\to}0$, whereas $Y_{n}$ is uniformly tight.
van der Vaart uses Prohorov's theorem (ii) to argue that there exists a subsequence $\{n_{j}\}$ along which $Y_{n_{j}}\overset{\mathcal{D}}{\to}Y$, for some $Y$, as $j\to\infty$.
The idea of van der Vaart comes from the fact that if $Y_{n_{j}}\overset{\mathcal{D}}{\to}Y$, then we can involve Slutsky's lemma to argue that $X_{n_{j}}Y_{n_{j}}\overset{\mathcal{D}}{\to}0\cdot Y=0$, and then use Prohorov's theorem (i) to show that $\{X_{n_{j}}Y_{n_{j}}\}$ is uniformly tight, thus bounded in probability: $X_{n_{j}}Y_{n_{j}}=O_{P}(1)$.
However, this proof does not seem complete. The proof only holds along this subsequence $\{n_{j}\}$ on which uniform tightness implies weak convergence, but what happens with the reamining terms of the sequence $\{n\}$?

van der Vaart, A. W., Asymptotic statistics, Cambridge Series in Statistical and Probabilistic Mathematics, 3. Cambridge: Cambridge Univ. Press. xv, 443 p. (1998). ZBL0910.62001..

 A: Following this post and this answer I think I got an argument in place.
Set $Z_{n}\doteq X_{n}Y_{n}$ to simplify the notation.
We have shown that any sequence $\{Z_{n}\}$ contains a subsequence $\{Z_{n_j}\}$ that converges weakly to zero, namely $Z_{n_j}\overset{\mathcal{D}}{\to}0$, and, hence, is uniformly tight by Prohorov's theorem (i). Furthermore, we can prove that weak convergence to a constant implies convergence in probability to the constant, namely $Z_{n_j}\overset{P}{\to}0$, see for example Theorem 2.7 in van der Vaart's Asymptotic statistics.
Given that $X_{n_j}=o_{P}(1)$ and $Y_{n_j}=O_{P}(1)$ we repeat van der Vaart's logic, Prohorov's theorem (ii) and Slutsky's lemma, to show that any subsequence $\{Z_{n_j}\}$ contains a further subsequence $\{Z_{n_{j_k}}\}$ which converges to weakly to zero.
Suppose now that $\{Z_{n}\}$ does not converge weakly to zero. In this case, there exist $\varepsilon>0$, $\delta>0$, and $\{n_j\}$ such that $P(\lVert Z_{n_j}\rVert>\varepsilon)\ge\delta$ for all $j$. However, $\{Z_{n_j}\}$ contains a further subsequence $\{Z_{n_{j_k}}\}$ that converges weakly to $0$ and consequently converges in probability to $0$. Hence, for all $j\in\{j_{k}\ |\ k\in\mathbb{N}\}$ we have $P(\lVert Z_{n_j}\rVert>\varepsilon)<\delta$. As this is a contradiction, $Z_{n}\overset{\mathcal{D}}{\to}0\implies Z_{n}\overset{P}{\to}0$.

van der Vaart, A. W., Asymptotic statistics, Cambridge Series in Statistical and Probabilistic Mathematics, 3. Cambridge: Cambridge Univ. Press. xv, 443 p. (1998). ZBL0910.62001.

A: The sequence indexed by the other {n} will also converge to zero, but that is not a helpful answer. Better is to prove the claim via a contradiction.
Suppose the whole sequence does not converge to zero. Then there is a subsequence that is bounded away from zero. But as you have discovered: this subsequence has further subsequence that does converge to zero. Contradiction.
By the way, a direct proof from the definitions is also possible, and to me more satisfying.
