 Solution

The union bound gives $\mathbb{P}({\bigcup }_{n\ge m}{A}_n)\le {\sum }_{n\ge m}\mathbb{P}(A_n)\to 0$. Taking the decreasing intersection proves the first Borel-Cantelli lemma.

If $X_n\to X$ almost surely, then $1_{\{|X_n-X|>\epsilon \}}\to 0$ almost surely; dominated convergence theorem gives convergence of its expectation, hence convergence in probability.

If convergence in probability holds, any subsequence has a further one with $\mathbb{P}(|X_{n(k_r)}-X|>2^{-r})<2^{-r}$. Borel-Cantelli then gives almost-sure convergence. Conversely, failure in probability supplies a subsequence with probabilities bounded below by some positive constant, and no further subsequence can converge almost surely because that would imply convergence in probability.