Define $S_n:=\sum_{j=1}^nX_j$ and $\varphi_n$ as the characteristic function of $S_n$. Using independence and normal distribution, we have
$$\varphi_n(t)=\exp\left(it\sum_{j=1}^n\mu_j\right)\cdot\exp\left(-\frac{t^2}2\sum_{j=1}^n\sigma_j^2\right).$$
As $S_n$ converges almost surely, we should have that $(\varphi_n(t))_{n\geqslant 1}$ is a convergent sequence for each $t$ to $\varphi(t)$, where $\varphi$ is a continuous function (the characteristic function of the limiting distribution) .
If $\sum_{j\geqslant 1}\sigma_j^2$ was divergent, then we would have $\varphi(t)=0$ if $t\neq 0$ and $\varphi(0)=1$, a contradiction. Define $s_n:=\sum_{j=1}^n\mu_j$. We have that for each $t$, the sequence $(e^{its_n})_{n\geqslant 1}$ is convergent hence $(s_n)_{n\geqslant 1}$ is convergent.
There is likely a proof somewhere on this site but I could not find it. Here I give a quick proof of my comment (since I originally mis-stated the result by forgetting the "lower bounded" restriction):
Let $\{X_i\}_{i=1}^{\infty}$ be a sequence of random variables, not necessarily identically distributed and not necessarily independent, that satisfy:
i) $E[X_i]=m_i$, where $m_i \in \mathbb{R}$ for all $i\in\{1, 2, 3, ...\}$.
ii) There is a constant $\sigma^2_{bound}$ such that $Var(X_i) \leq \sigma^2_{bound}$ for all $i \in \{1, 2, 3, ...\}$.
iii) The variables are pairwise uncorrelated, so $E[(X_i-m_i)(X_j-m_j)]=0$ for all $i \neq j$.
iv) There is a value $b \in \mathbb{R}$ such that, with prob 1, $X_i-m_i\geq b$ for all $i \in \{1, 2, 3, ...\}$.
Define $L_n = \frac{1}{n}\sum_{i=1}^n (X_i-m_i)$. Then $L_n\rightarrow 0$ with prob 1.
Proof: Since the variables are pairwise uncorrelated with bounded variance, we easily find for all $n$:
$$ E[L_n^2] = \frac{1}{n^2}\sum_{i=1}^n \sigma_i^2 \leq \frac{\sigma_{bound}^2}{n} $$
Fix $\epsilon>0$. It follows that:
$$ P[|L_n|>\epsilon] = P[L_n^2 \leq \epsilon^2] \leq \frac{E[L_n^2]}{\epsilon^2} \leq \frac{\sigma_{bound}^2}{n\epsilon^2} $$
Hence:
$$ \sum_{n=1}^{\infty} P[|L_{n^2}|>\epsilon] \leq \sum_{n=1}^{\infty}\frac{\sigma_{bound}^2}{n^2\epsilon^2} < \infty $$
and so $L_{n^2}\rightarrow 0$ with probability 1 by the Borel-Cantelli Lemma. That is, the $L_n$ values converge over the sparse subsequence $n\in\{1, 4, 9, 16, ...\}$.
Since $L_n \geq b$ for all $n$ and $L_{n^2}\rightarrow 0$ with probability 1, it can be shown that $L_n\rightarrow 0$ with probability 1. $\Box$
The lower bounded condition is typically treated by writing $X_n = X_n^+ - X_n^-$ where $X_n^+$ and $X_n^-$ are nonnegative and defined $X_n^+=\max[X_n,0]$, $X_n^-=-\min[X_n,0]$. If $X_n$ and $X_i$ are independent, then $X_n^+$ and $X_i^+$ are also independent. So the lower bounded condition can be removed for the case when variables are independent. However, if $X_n$ and $X_i$ are uncorrelated, that does not mean $X_n^+$ and $X_i^+$ are uncorrelated. So it is not clear to me if the lower-bounded condition can be removed when "independence" is replaced by the weaker condition "pairwise uncorrelated."
Best Answer
We can use a symmetrization trick: let $(X'_j)_{j \geqslant 1}$ and $(X''_j)_{j \geqslant 1}$ be independent copies of the sequence $(X_j)_{j\geqslant 1}$. Using the Lindeberg central limit theorem and exploiting the fact that $|X'_j-X''_j|\leqslant 2M$, we can show that for each positive $A$, $$\lim_{n\to\infty}\mathbb P\left\{\left|\sum_{j=1}^n X'_j-X''_j \right| \leqslant A \right\} =0.$$ To conclude, notice that for each $A\gt 0$, the inclusion $$\left\{\left|\sum_{j=1}^n X'_j \right| \leqslant \frac A2 \right\}\cap\left\{\left|\sum_{j=1}^n X''_j \right| \leqslant \frac A2 \right\}\subset \left\{\left|\sum_{j=1}^n X'_j-X''_j \right| \leqslant A \right\}$$ holds, and the events in the left hand side are independent and their common probability is $\mathbb P\left\{\left|\sum_{j=1}^n X_j \right| \leqslant \frac A2 \right\}$.