Suppose $X_n\to X$ in probability. Then $X_n \to X$ almost surely if and only if every subsequence has a further subsequence converging almost surely to $X$. However, given any subsequence $X_{n_k}$ we have $X_{n_k}\to X$ in probability so that $X_{n_k}$ has a subsequence converging to $X$ almost surely. Hence $X_n\to X$ almost surely.
Where is the error?
The claim "$X_n \to X$ almost surely if and only if every subsequence has a further subsequence converging almost surely to $X$" is false.
You might have made this error since in a topological space it is true that a sequence converges to $x$ if and only if every subsequence has a further subsequence converging to $x$. However, convergence almost surely is not a mode of convergence associated to a topology.
To see what goes wrong, let me write a failed proof of the (false) claim. For real valued random variables, you might have tried to mimic the usual proof of the topological fact by writing something like:
If $X_n \not \to X$ almost surely then there exists a set $\Omega' \subseteq \Omega$ of positive measure such that for all $\omega \in \Omega'$, $X_n(\omega) \not \to X(\omega).$ Then for each $\omega$, there exists a subsequence $X_{n_k(\omega)}(\omega)$ and $\varepsilon(\omega) > 0$ such that $|X_{n_k(\omega)} - X(\omega)| > \varepsilon(\omega)$. At this point, you get stuck. The subsequence depends on $\omega$, but you only know that there is a subsequence of $n_k(\omega)$, $n_{k_j}(\omega)$ such that $X_{n_{k_j}(\omega)}$ converges almost surely. You do not know that $X_{n_{k_j}(\omega)}(\omega) \to X(\omega)$ which is what you would need.