I'm trying to prove this intuitive result. Could you have a check on my attempt?
Let $(X, d)$ be a metric space. Let $\mathcal{M} :=\mathcal{M}(X)$ and $\mathcal{P} :=\mathcal{P}(X)$ be the sets all non-negative finite Borel measures and the set all Borel probability measures on $X$ respectively. Let $d_P$ be the Prokhorov metric on $\mathcal{M}$. Then $\mathcal P$ is closed in $\mathcal M$.
I post my proof separately as below answer. If other people post an answer, of course I will happily accept theirs. Otherwise, this allows me to subsequently remove this question from unanswered list.
Proof: Assume $\mu, \mu_1, \mu_2,\ldots \in \mathcal M$ such that $\mu_n \to \mu$ in $d_P$. Let $\alpha_n := d_P(\mu_n, \mu) + 1/n$ and $C$ be closed in $X$. Let $C_{n} := \{x\in X \mid d(x, C) < \alpha_n\}$. Then $\alpha_n \to 0$ and $C_n \searrow C$. By definition of $d_P$, we have $$ \mu_n (C) \le \mu(C_n) + \alpha_n \quad \forall n. $$
Then $$ \limsup_n \mu_n(C) \le \limsup_n [\mu(C_n) + \alpha_n] = \limsup_n \mu(C_n) = \mu(C). $$
It follows that $\mu_n \to \mu$ weakly.
Let $\mu_1, \mu_2, \ldots \in \mathcal P$ and $\mu \in \mathcal M$ such that $\mu_n \to \mu$ in $d_P$. Notice that $\partial X = \overline X \setminus \mathring X = \emptyset$, so $\mu (\partial X)=0$. By our Lemma, $\mu_n \to \mu$ weakly. So $\mu_n (X) \to \mu(X)$. Clearly, $\mu_n(X)=1$ for all $n$. Hence $\mu(X)=1$ and thus $\mu \in \mathcal P$. This completes the proof.