Suppose $X_n\overset{p}{\to} X$. Does this imply $\Pr(X_n=X)\to 1$ as $n\to\infty$?
My approach would be to prove the equivalent statement $\Pr(X_n\neq X)\to 0$ as $n\to\infty$. Since $\big\{ X_n\neq X \big\}=\bigcup\limits_{k=1}^{\infty} \Big\{\lvert X_n-X \lvert > \frac{1}{k} \Big\} $, we have $\Pr(X_n\neq X)\leq \sum\limits_{k=1}^{\infty}\Pr(\lvert X_n-X \lvert > \frac{1}{k})$ by countable subadditivity. Because we assumed $X_n\overset{p}{\to} X$, each term inside the sum can be made arbitrarily small if $n$ is chosen large enough. Am having trouble with filling the details. Any help is greatly appreciated.
Let $X_n = 1/n$ and $X=0$. Then for any $\epsilon > 0$, the probability $P(|X_n - X| > \epsilon) = \mathbf{1}_{\{1/n > \epsilon\}}$ is zero for all sufficiently large $n$, so $X_n \overset{p}{\to} X$. However, $P(X_n = X) = 0$ for all $n$.
Regarding your proof attempt: it is true that each term in the sum can be made arbitrarily small if $n$ is large enough, but it may not be true that you can find an $n$ that simultaneously makes all terms small enough for the entire sum to be to be small.