Conclude that $\frac{1}N\sum_{n=1}^{N}X_n\to 0$ almost surely (without using strong law of large numbers) Given a sequence of IID random variables with expectation zero and finite variance, $X_1,X_2,...$

Prove that $S_N=\sum_{n=1}^{N}\frac{X_n}{n}$ is a martingale, and
  conclude that $\frac{1}N\sum_{n=1}^{N}X_n\to 0$ almost surely using
  this exercise

So for showing that $\sum_{n=1}^{N}\frac{X_n}{n}$ is a martingale I think I got it right:
$E[S_{N}|S_1,...,S_{N-1}]=E[\frac{X_N}{N}+S_{N-1}|S_1,...,S_{N-1}]=E[\frac{X_N}{N}|S_1,...,S_{N-1}]+E[S_{N-1}|S_1,...,S_{N-1}]=\frac{1}{N}E[X_N]+S_{N-1}=0+S_{N-1}=S_{N-1}$
For the second part  though I'm not sure, can anyone help me?
 A: The result follows immediately by the strong law of large numbers. The law states that for a sequence of iid random variables with finite mean,
\begin{align*}
P(\lim_n \bar{X}_n = EX) = 1
\end{align*}
i.e. the empirical mean $\frac{1}{N} \sum_{i=1}^N X_i$ converges almost surely to $EX=0$.
To show it using the linked exercise, we invoke the Martingale Convergence Theorem on $\{S_n\}$ so that $S_N = \sum_{n=1}^N \frac{X_n}{n}$ converges. Then by the linked exercise, the empirical average converges to 0.
To show the sufficient condition for MCT, namely that $\sup_N E|S_N| < \infty$, observe that
\begin{align*}
Var|S_N| &= Var\Big|\sum_{n=1}^N \frac{X_n}{n} \Big| \\
&\leq \sum_{n=1}^N \frac{1}{n^2}Var|X_n| \\
&= Var|X_1| \sum_{n=1}^{N} \frac{1}{n^2} < \infty
\end{align*}
by assumption. Since $Var|S_N| = E(S_N^2) - (ES_N)^2 = E(S_N^2) - 0 = E(S_N^2)$, we have that the measurable function $S_N$ is in $L^2$. Since the probability measure is finite, $S_N$ is also an element of $L^1$. Thus, $E|S_N| < \infty$.
