Prove Kolmogorov's zero one law using martingales I am supposed to provide a martingale proof of Kolmogorov's zero-one law.

Hint Let $X_n$ be independent random variables and let $\mathcal C_\infty$ be the corresponding tail $\sigma$-algebra. Let $C \in
 \mathcal C_\infty$ and $\mathcal F_n = \sigma(X_j; 0 \le j \le n)$.
Show that $E[1_C\mid \mathcal F_n] = P(C)$. Then show that $\lim_{n
 \to \infty} E[1_c \mid \mathcal F_n] = 1_C$ almost surely and deduce
  that $P(C) = 0$ or $1$

I have no idea how to approach this. I would like an answer but I would love an answer that provides insights about all this. I don't have any intuition about what $C_\infty$ is, and why should I follow the outlined proof, and what's the role of martingales.
EDIT 1
Let me try to provide a proof. Please point out any mistake, inaccuracy or redundancy that you can spot! I'll be most grateful.
Let us note that for every finite $N$, $\mathcal C_\infty$ is independent with $\mathcal F_N$. In fact $$\mathcal C_\infty = \bigcap_{n=1}^\infty \sigma\left(\bigcup_{m \ge n} F_m\right) = \bigcap_{n=N+1}^\infty \sigma\left(\bigcup_{m \ge n} F_m\right)$$
So $\mathcal C_\infty$ is the result of operations applied only on sigma algebras independent with $\mathcal F_N$
This implies $E[1_c \mid \mathcal F_n] = E[1_C] = P(C)$.
Let $\mathcal B_n = \sigma\left(\bigcup_{i \le n} \mathcal F_i\right)$. By the exact same reasoning, we find $E[1_c \mid \mathcal B_n] = P(C) \implies E[1_c \mid \mathcal B_n] = E[1_c \mid \mathcal F_n]$
This was necessary because now $\mathcal B_n$ is an increasing sequence of sigma algebra and I can apply Levy's zero one law:
$$\lim_{n \to \infty} E[1_C \mid \mathcal F_n] = \lim_{n\to \infty} E[1_C \mid \mathcal B_n] =  E[1_C \mid \mathcal B_\infty]\text{    a.s.}$$
Where $\mathcal B_\infty = \sigma\left(\bigcup_{n \ge 0} \mathcal B_n\right)$
Since $\mathcal C_\infty = \inf \mathcal B_\infty \implies \mathcal C_\infty \subset \mathcal B_\infty$ (is this correct?) the fact that $1_C$ is measurable with respect to $\mathcal C_\infty$ implies that $1_C$ is measurable with respect to $\mathcal B_\infty$, so
$$\lim_{n \to \infty} E[1_C \mid \mathcal F_n] = E[1_C \mid \mathcal B_\infty] = 1_C$$
Now $\lim_{n \to \infty} E[1_C \mid \mathcal F_n] = P(C) = 1_C$, hence $P(C) = 0$ or $1$.
Remarks
The proof seems somewhat convoluted and it is probably better done in another way. Also, I am not really sure is it correct.
Moreover, I did not use directly martingale properties; I know Levy's zero one law is a consequence of the Martingale Convergence theorem, though.
EDIT 2
I deleted my answer and decided to put my attempt in the question, so I can award points to anyone who is kind enough to help me. I didn't do so previously because the wall of text may be a deterrent to read the whole question.
 A: First of all (and most importantly): The idea of your proof is correct. However, there are some things which can be improved:
$(\mathcal{F}_n)_{n \in \mathbb{N}}$ is a filtration, i.e. $\mathcal{F}_n$ is a $\sigma$-algebra for each $n \in \mathbb{N}$ and $\mathcal{F}_n \subseteq \mathcal{F}_{n+1}$. This follows directly from the definition of $\mathcal{F}_n$. This means in particular that you can apply Lévy's zero one law directly to the sequence
$$\mathbb{E}(1_C \mid \mathcal{F}_n);$$
there is no need for $\mathcal{B}_n$. (In fact, since the $\sigma$-algebras $\mathcal{F}_n$ are increasing, it holds that $\mathcal{B}_n = \mathcal{F}_n$.) This makes the proof much more easier.
Concerning martingale properties: You are right; Lévy's zero one law is a direct conseqeunce of the martingale convergence theorem. The martingale you are considering in this case is actually
$$X_n := \mathbb{E}(1_C \mid \mathcal{F}_{n}).$$
If it is not clear to you why this is a martingale, then try to prove it - it is a good exericse. :)
