I know that for a general random variable $X$ (defined on some probability space) it holds true that $\mathbb{P}(X<\infty)=1$ implies that $\mathbb{E}(X)<\infty$ (and I already know how to prove this).
On the other hand, if \begin{equation*} \mathbb{E}(X)<\infty \end{equation*} does it follow that \begin{equation*} \mathbb{P}(X<\infty)=1 \end{equation*} (i.e. $X<\infty$ a.s.)? If so, how can one prove that?
Actually, it is not true that $P(X<\infty)=1$ implies $E(X)<\infty$, and you should find a counterexample to that assertion. On the other hand, the converse is true. The reason is: If $P(|X|=\infty)>0$ then $$E(|X|)\ge E(|X|I(|X|=\infty))=\infty\cdot P(|X|=\infty)=\infty.$$
Another proof uses Markov's inequality. If $E(|X|)<\infty$ then $$P(|X|>M)\le \frac{E|X|}M $$ so by continuity from above $$P(|X|=\infty) = P\left(\bigcap_M \{|X|>M\}\right)=\lim_M P(|X|>M)=0.$$