Is convergence in probability equivalent to convergence of all conditioned distributions? I have a simple question regarding potentially an alternative characterisation of convergence in probability.

Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space, and for each $E \in \mathcal{F}$ with $\mathbb{P}(E)>0$, define the probability measure $\mathbb{P}_E$ on $(\Omega,\mathcal{F})$ by $\mathbb{P}_E(A)=\frac{\mathbb{P}(E \cap A)}{\mathbb{P}(E)}$.
Let $(S,d)$ be a separable metric space, and suppose we have a measurable function $X_n \colon \Omega \to S$ for each $n \in \mathbb{N} \cup \{\infty\}$.


Suppose that for every $E \in \mathcal{F}$ with $\mathbb{P}(E)>0$, $X_n \to X_\infty$ in distribution over $(\Omega,\mathcal{F},\mathbb{P}_E)$ as $n \to \infty$. Does it follow that $X_n \to X_\infty$ in probability over $(\Omega,\mathcal{F},\mathbb{P})$ as $n \to \infty$?


[The converse direction is immediate from the fact that convergence in probability implies convergence in distribution, since convergence in probability over $(\Omega,\mathcal{F},\mathbb{P})$ clearly implies convergence in probability over $(\Omega,\mathcal{F},\mathbb{P}_E)$.]
 A: I've managed to work out now that the answer is yes.
Throughout the following, we assume as in the question that $(S,d)$ is a separable metric space.
Definition. For any Borel probability measure $\mu$ on $S$ and any $\varepsilon>0$, a $(\mu,\varepsilon)$-partition of $S$ is a disjoint collection $\mathcal{P}$ of open sets $U \subset S$ such that $\mu(\bigcup_{U\in\mathcal{P}}U)=1$ and for each $U \in \mathcal{P}$, $\mathrm{diam}(U) \leq \varepsilon$ and $\mu(U)>0$.
Lemma. For any Borel probability measure $\mu$ on $S$ and any $\varepsilon>0$, there exists a $(\mu,\varepsilon)$-partition of $S$.
[The proof is based on what I think is a standard trick for constructing sets whose boundary has zero measure.]
Proof of Lemma. Let $(x_n)_{n\geq 1}$ be a dense sequence in $S$. For each $n$, since the uncountable family $\{\{x \in S : d(x,x_n)=r\}: r \in [\frac{\varepsilon}{3},\frac{\varepsilon}{2}]\}$ is mutually disjoint, there must exist $r_n \in [\frac{\varepsilon}{3},\frac{\varepsilon}{2}]$ such that $\mu(x \in S : d(x,x_n)=r_n)=0$. Now let $A_1=B_{r_1}(x_1)$ and for $n \geq 2$ let $A_n=B_{r_n}(x_n) \setminus \bigcup_{i=1}^{n-1} B_{r_i}(x_i)$. Note that $\{A_n:n \geq 1\}$ forms a partition of $S$ (in the purely set-theoretic sense). Since we have the general facts that $\partial(A \cup B) \subset (\partial A) \cup (\partial B)$ and $\partial(A \setminus B) \subset (\partial A) \cup (\partial B)$, it follows in particular that $\partial A_n$ is a $\mu$-null set for each $n$. So let $\mathcal{P}'=\{A_n^\circ: n \geq 1\}$ and let $\mathcal{P}$ be the set of all $\mu$-positive-measure members of $\mathcal{P}'$.
Proof that the answer to the question is yes. Fix $\varepsilon>0$; we need to show that $\mathbb{P}(d(X_n,X_\infty)\leq\varepsilon) \to 1$ as $n \to \infty$. Let $\mu$ be the distribution of $X_\infty$, and let $\mathcal{P}$ be a $(\mu,\varepsilon)$-partition of $S$. For each $U \in \mathcal{P}$, letting $E_U$ be the event $\{X_\infty \in U\}$, we obviously have that $\mathbb{P}_{E_U}(X_\infty \in U)=1$; and so since $U$ is open and $X_n$ converges in distribution to $X_\infty$ over $(\Omega,\mathcal{F},\mathbb{P}_{E_U})$, it follows that $\mathbb{P}_{E_U}(X_n \in U) \to 1$ as $n \to \infty$. In other words,
$$ \mathbb{P}(X_n \in U \textrm{ and } X_\infty \in U ) \to \mathbb{P}(X_\infty \in U) \textrm{ as } n\to\infty $$
for each $U \in \mathcal{P}$. Therefore, by the discrete dominated convergence theorem we can sum over all $U \in \mathcal{P}$ to give
$$ \mathbb{P}(\exists \ U \in \mathcal{P} \textrm{ s.t. } X_n \in U \textrm{ and } X_\infty \in U ) \to 1 \textrm{ as } n\to\infty. $$
But for each $n$, we have that
$$ \{ \exists \ U \in \mathcal{P} \textrm{ s.t. } X_n \in U \textrm{ and } X_\infty \in U \} \subset \{d(X_n,X_\infty) \leq \varepsilon\}, $$
and so it follows that $\mathbb{P}(d(X_n,X_\infty)\leq\varepsilon) \to 1$ as $n \to \infty$ as required.
