First of all, note that you have to show
$$\sum_{n \geq 1} \mathbb{P}(|X_n| \geq \sqrt{n}) = \color{red}{\infty} \tag{1}$$
in order to apply the Borel-Cantelli-lemma.
Suppose that $(1)$ does not hold true, i.e. $\sum_{n \geq 1} \mathbb{P}(|X_n| \geq \sqrt{n}) < \infty$. Since the random variables are identically distributed, this implies
$$\sum_{n \geq 1} \mathbb{P}(|X_1|^2 \geq n)< \infty. \tag{2}$$
Now recall that for any non-negative random variable $Y$ we have
$$\mathbb{E}(Y) \leq \sum_{n \geq 0} \mathbb{P}(Y \geq n)$$
(this is a direct consequence of the pointwise inequality $Y \leq \sum_{n \geq 0} 1_{\{Y \geq n\}}$). Applying this for $Y= |X_1|^2$, we conclude from $(2)$ that
$$\mathbb{E}(X_1^2) \leq \sum_{n \geq 0} \mathbb{P}(|X_1|^2 \geq n)<\infty$$
in contradiction to our assumption that $X_1$ has infinite variance.
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
$S_{n}=n\bar{X}_{n}$ where $\bar{X}_{n}:=\frac{1}{n}\left(X_{1}+\cdots+X_{n}\right)$ and $P\left(\bar{X}_{n}\to\mu\right)=1$ according to the strong law of large numbers.
If $\bar{X}_{n}\left(\omega\right)\to\mu>0$ then $S_n(\omega)=n\bar{X}_{n}\left(\omega\right)\rightarrow+\infty$.
So: $$\left\{ \bar{X}_{n}\to\mu\right\} \subseteq\left\{ S_{n}\to+\infty\right\}$$ and consequently $$P(S_n\to+\infty)\geq P(\bar{X}_{n}\to\mu)=1$$