Sorry to unearth the topic, but I've got something here that might interest people, linked with this thread.
I ran across the following variant of your problem in T. Cacoullos Exercices in probability (Springer, 1989), exercice 254 : he calls it "theorem of Barnstein" (sic), but I couldn't find any clue about who this Barnstein is ; if it's a typo, then I don't know any variant of WLLN by Bernstein. Here's the statement :
Let $X_1, ..., X_n, ...$ be centered random variables.
If there exist a constant $c>0$ such that for every, $i$,
$\mathbf{V}[X_i] \leq c$ and if the following condition holds :
$$\lim_{|i-j| \to +\infty} \mathrm{Cov}(X_i, X_j) = 0$$ Then, the weak
law of large numbers hold.
This is a small generalization of your problem. The proof is very similar and consists in bounding the variance of $S_n /n$ in order to conclude with Chebyshev. To bound the variance, here's the argument (everything is very similar to what you wrote).
First, note that by Cauchy-Schwarz, $|\mathrm{Cov}(X_i, X_j)| \leq c$. Therefore, noting $S_n = X_1 + ... + X_n$,
$$\mathbf{V}[S_n] \leq \sum_{i=1}^{n} c + 2\sum_{i=1}^n \sum_{j=i+1}^{n}\mathrm{Cov}(X_i, X_j)$$
Choose $\epsilon >0$, take $N$ such that $\forall |i-j|>N$, we have $\mathrm{Cov}(X_i, X_j) < \epsilon$. Now if $n$ is greater than $N$ (so we don't have problems with the indexes) split the sum over $j$ before and after $N$, so we have
$$\sum_{i=1}^n \sum_{j=1}^{i-1}\mathrm{Cov}(X_i, X_j) = \sum_{i=1}^n \sum_{j=i+1}^{i+N}\mathrm{Cov}(X_i, X_j) + \sum_{i=1}^n \sum_{j=i+N+1}^{n}\mathrm{Cov}(X_i, X_j) $$
Invoking triangle inequality, we have
$$\left|\sum_{i=1}^n \sum_{j=1}^{i-1}\mathrm{Cov}(X_i, X_j) \right| \leq \sum_{i=1}^n Nc + \sum_{i=1}^n \sum_{j=i+N+1}^{n}\epsilon \leq nNc + n^2 \epsilon $$
Therefore,
$$\mathbf{V}[S_n /n] \leq \frac{c}{n} + \frac{2Nc}{n} + \epsilon $$
This clearly proves that $\mathbf{V}[S_n /n] \to 0$ as $n \to + \infty$, ending the proof of the mysteriously so-called "theorem of Barnstein". I hope this will help someone !