Convergence in probability implies a.s. convergence in a countable space

probabilityprobability theoryprobability-limit-theorems

Let be $$(\Omega, \mathcal{F}, \mathcal{P})$$ a probability space, $$(X_{n})$$ a sequence of randiom variables and $$X$$ a random variable. Let be $$\Omega$$ countable and $$\mathcal{F}$$ a power set of $$\Omega$$.

Show that $$X_n\xrightarrow{p} X, n \rightarrow \infty \ \ \implies X_n \xrightarrow{a.s.} X, n \rightarrow \infty$$

I have done some research and I have found out that the statement is not true in general but I have no idea how to prove it for this special case. I appreciate your help in advance a lot.

Let $$F=\{\lim_{n\to\infty}X_n=X\}$$ and suppose that $$\mathsf{P}(F)<1$$. Since $$F^c$$ is countable, there exists $$\omega'\in F^c$$ such that $$\mathsf{P}(\{\omega'\})=p>0$$. Thus, there exists a subsequence $$n_k$$ such that $$|X_{n_k}(\omega')-X(\omega')|>\epsilon>0$$. However, this implies $$\mathsf{P}(|X_{n_k}-X|>\epsilon)\ge \mathsf{P}(\{\omega'\})=p\not\to 0.$$