Sample variance converge almost surely

Since you used an inequality to obtain an estimate for $\mathbb{P}(|S_n-\mathbb{E}S_n|>\varepsilon)$ the fact that the sum does not converge does not imply that $S_n$ does not converge almost surely.

It's easier like that:

$$\begin{align} (X_i-\mu+(\mu-\bar{X}))^2 &= (X_i-\mu)^2 + 2 (X_i-\mu) \cdot (\mu-\bar{X})+(\mu-\bar{X})^2 \\ \Rightarrow S_n= \frac{1}{n} \sum_{i=1}^n (X_i-\bar{X})^2 &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 + 2 (\mu-\bar{X}) \underbrace{\frac{1}{n} \cdot \sum_{i=1}^n (X_i-\mu)}_{\left(\frac{1}{n} \sum_{i=1}^n X_i\right)-\mu =(\bar{X}-\mu)} + (\mu-\bar{X})^2 \\ &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 - (\bar{X}-\mu)^2 \end{align}$$

By the strong law of large numbers we obtain

$$\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \to \mu \quad \text{almost surely} \\ \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 \to \mathbb{E} \left( (X_i-\mu)^2 \right)=\sigma^2 \quad \text{almost surely} $$

Hence $S_n \to \sigma^2$ almost surely.

This is much after 2012, but since I chanced upon this, I'd like to just offer an alternative solution (but similar in spirit):

$$\frac{1}{n} \sum_{i=1}^n (X_i-\bar{X})^2 = \frac{1}{n}\sum_{i=1}^n (X_i^2-2X_i\bar{X}+(\bar{X})^2)=\frac{\sum_{i=1}^n X_i^2}{n}-\left(\frac{\sum_{i=1}^n X_i}{n}\right)^2$$

Since $X_i^2$ are i.i.d. with finite mean $\mu^2 + \sigma^2$, by strong law of large numbers (SLLN), $\frac{\sum_{i=1}^n X_i^2}{n} \to \mu^2 + \sigma^2$ almost surely.

In addition, SLLN says $\frac{\sum_{i=1}^n X_i}{n} \to \mu$ almost surely. By Continuous Mapping Theorem (or direct reasoning), $\left(\frac{\sum_{i=1}^n X_i}{n}\right)^2 \to \mu^2$.

Hence, $S_n \to \sigma^2$ almost surely.