Let $(X_i)_{i \geq 1}$ i.i.d. and $\mathbb{E}(|X_1|)< \infty$, w.l.o.g. $\mathbb{E}(X_1) = 0$. This in turn implies: $$ \sum_{n=1}^{\infty} \mathbb{P}(|X_1| \geq n) \leq \int_0^{\infty} \mathbb{P}(|X_1| \geq t) \ dt = \mathbb{E} (|X_1|)< \infty $$
Now my professor argues that by Borel-Cantelli, a.s. $|X_n| \leq n$ for $n$ large enough, thus we have, except on a set of probability zero, $X_n \chi_{|X_n|\leq n}=X_n$ (where $\chi_{|X_n|\leq n}$ is the characteristic function on $\{\omega: |X_n(\omega)|\leq n\}$). Then the proof of the SLLN continues.
I tried to formalise this: By B.C. I.:
\begin{equation} \mathbb{P}\left(\lim \inf_{n \rightarrow \infty} \{\omega \ : \ |X_n(\omega)| < n\}\right)=\mathbb{P}\left(\bigcup_{N=1}^{\infty} \bigcap_{n=N}^{\infty} \{\omega \ : \ |X_n(\omega)| < n\}\right)=1 \end{equation}
I.e. \begin{equation} \mathbb{P}(\{\omega \ | \ \exists N \ \forall n \geq N: \ |X_n(\omega)| < n \})=1 \end{equation}
How does my professor come to the conclusion: $$ \exists N \ \forall n \geq N:\mathbb{P}(\{\omega \ | \ |X_n(\omega)| < n \}) = 1 $$
This is a way stronger statement, isn’t it? Or is he saying something different?
Until \begin{equation} \mathbb{P}(\{\omega \ | \ \exists N \ \forall n \geq N: \ |X_n(\omega)| < n \})=1 \end{equation} it is correct. It means that almost surely, for $\omega\in\Omega$ there exists $N_\omega\in\mathbb N$ such that $\vert X_n(\omega)\vert<n$ for all integers $n\ge N_\omega$. It is this fact that you then use in the proof. You did not write the dependence of $N$ in $\omega$, and wrote simply $N$ instead of $N_\omega$, but that is for convenience, not because the dependence does not exist.
Of course you cannot deduce that $$ \exists N \ \forall n \geq N:\mathbb{P}(\{\omega \ | \ |X_n(\omega)| < n \}) = 1, $$ which tells you the same as above except that you can choose a universal $N$ which does not depend on the choice of $\omega$, which is a much stronger statement. More convenient, but not needed.