Consider a sequence of real-valued random variables, $\{X_n\}_{n}$.
Suppose that
(*) $X_n \rightarrow_p X$.
Consider
$\lim_{n \rightarrow \infty} P(X_n>0)$.
What I have seen doing in several books is the following: since $X_n\rightarrow_p X$ then
(**) $\lim_{n \rightarrow \infty} P(X_n>0)=P(X>0)$
Could you indicate step by step how (*) implies (**)?
Generalizing to $c$ in place of zero, suppose $c$ is a continuity point of the distribution of $X$ (the result is not necessarily true otherwise: try $X_n=c+1/n$, $X=c$; then $P(X_n>c)=1>0=P(X>c)$).
Suppose that $X_n\le c$. Let $\epsilon>0$. Since $X_n\to X$ in probability, the intuition is to consider two cases: either $|X_n-X|>\epsilon$ or $|X_n-X|\le\epsilon$. The first case has small probability. In the second case, we know for sure that $X\le c+\epsilon$. This leads to the inequality $$ P(X_n\le c)\le P(|X_n-X|>\epsilon) + P(X\le c+\epsilon),\tag1 $$ which holds for all $n$ and $\epsilon$. You see where this is going: As $n\to\infty$, the RHS of (1) tends to $P(X\le c+\epsilon)$. But this holds for every $\epsilon$, so in the limit the upper bound should be $P(X\le c)$. More formally, you could apply $\limsup_n$ to both sides of (1), to obtain $$\limsup_nP(X_n\le c)\le P(X\le c+\epsilon) $$ for every $\epsilon$, and then invoke continuity of the distribution of $X$ at $c$, to deduce $$ \limsup_nP(X_n\le c)\le P(X\le c).\tag2 $$ Similarly start with $$ P(X\le c-\epsilon)\le P(|X_n-X|>\epsilon) + P(X_n\le c),\tag3 $$ and conclude $$ P(X\le c)\le \liminf_nP(X_n\le c).\tag4 $$ But $\liminf\le\limsup$ for any sequence, so (2) and (4) imply that $P(X_n\le c)$ converges and $$ \lim_n P(X_n\le c)=P(X\le c). $$ (Note that we apply $\limsup$ to (1) but $\liminf$ to (3). Why not the other way around?)