Why can we apply the Kolmogorov's zero one law for $\{ \text{ lim }X_n \leq x \}$

Let $X_n$ be a sequence of independent random variables that converges almost surely to $X$. Show that $X$ is constant almost surely.

Now, the way this was solved by the teacher, was by using the Kolmogorov's 0-1 law. That is:

$\{ X \leq x \} = \{ \underset{n \infty}{\lim}X_n \leq x \}$ Thus by Kolmogorov's 0-1 law, the probability of this set is either $0$ or $1$.

He then proceeds to use this to prove the statement. Now, my issue with this is that in order to work use this theorem, we have to introduce sigma algebras generated by $X_i$. I guess he did it implicitly, but I don't see how. Can someone show me how it would be done with the sigma algebras?

I am asking here on MSE, as I have no more lectures and no access to my teachers.

If $X_n$ are independent and $${\cal T}_N = \sigma\big(\{X_n\;:\; n \ge N\}\big) \\ {\cal T} = \bigcap_{N=1}^\infty {\cal T}_N$$ then every event in $\cal T$ has either probability $0$ or $1$.
So: show the event $\{\;X \le x \:\}$ belongs to $\cal T$.