Note that $$ \frac{S_{n+1}}{n+1}-\frac{S_n}n=-\frac{S_n}{n(n+1)}+\frac{X_{n+1}}{n+1},$$
hence the sequence $(X_n/n)_{n\geqslant 1}$ converges to $0$ almost surely.
The series $\sum_n \mathbb P(|X_n|\geqslant n) $ is convergent; otherwise, by the second Borel-Cantelli's lemma, we would have $\mathbb P( \limsup_n\{|X_n|\geqslant n\})=1$, which contradicts the almost sure convergence to $0$ of the sequence $(X_n/n)_{n\geqslant 1}$.
This answer is based on a coupling argument, which makes rigorous the comment by LJG in the answer by Davide Giraudo. We will assume the (usual) strong law of large numbers for i.i.d. sequences, but nothing else.
Namely, let $X_n$ be independent random variables with $X_n \sim Bern(p_n)$ where $p_n \downarrow p$. Define a sequence $Y_n$ of i.i.d. Bernoulli($p$) random variables as follows. If $X_n=0$, let $Y_n=0$. If $X_n=1$, flip an (independent, unfair) coin, whose probability of heads is $1-p/p_n$. If heads, let $Y_n=0$. If tails let $Y_n=1$. Clearly $Y_n \leq X_n$ for all $n$, and the $Y_n$ are i.i.d. Bern($p$). By the strong law of large numbers $$\liminf_{n \to \infty}\frac{1}{n} \sum_1^n X_k \geq \liminf_{n \to \infty}\frac{1}{n} \sum_1^n Y_k =p, \;\;\;\;\;\;\;a.s.$$
Now let $q>p$. For large enough $n$ we have that $p_n<q$. Since the limit of Cesaro means does not depend on the omission of some finite number of terms, we may assume wlog that $p_n<q$ for all $n$. We now define a sequence $Z_n$ of i.i.d. Bernoulli($q$) random variables as follows. If $X_n=1$, let $Z_n=1$. If $X_n=0$, flip an (independent, unfair) coin, whose probability of heads is $(q-p_n)/(1-p_n)$. If heads, let $Z_n=1$. If tails let $Z_n=0$. Clearly $Z_n \geq X_n$ for all $n$, and the $Z_n$ are i.i.d. Bern($q$). By the strong law of large numbers, $$\limsup_{n \to \infty}\frac{1}{n} \sum_1^n X_k \leq \limsup_{n \to \infty}\frac{1}{n} \sum_1^n Z_k =q, \;\;\;\;\;\;\;a.s.$$ But since $q>p$ was arbitrary, we conclude (after taking the intersection over countably many $q$'s converging down to $p$) that $$\limsup_{n \to \infty}\frac{1}{n} \sum_1^n X_k \leq p, \;\;\;\;a.s.$$ which gives the result.
Best Answer
As noted by @Michael, Etemadi's theorem assumes the variables are identically distributed, so the question is to show this assumption cannot be dropped. Suppose that for $k \ge 1$, the independent variables $Y_k$ take the values $0,1$, with $P(Y_k=1)=(k\log(1+ k))^{-1}$. Let $X_k=k\log(1+ k) Y_k$. By the Borel-Cantelli lemma, the event $Y_k=1$ happens infinitely often almost surely. For each $k$ such that $Y_k=1$ we have $X_k/k=\log(1+k)$, so the limsup mentioned in the original problem is indeed $\infty$ with probability 1.