I am trying to show that the sample variance converges to the population variance in using the Weak Law of Large Numbers
$$\begin{align} \\ \Rightarrow S_n= \frac{1}{n} \sum_{i=1}^n (X_i-\bar{X})^2 &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 + 2 (\mu-\bar{X}) \underbrace{\frac{1}{n} \cdot \sum_{i=1}^n (X_i-\mu)}_{\left(\frac{1}{n} \sum_{i=1}^n X_i\right)-\mu =(\bar{X}-\mu)} + (\mu-\bar{X})^2 \\ &= \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 - (\bar{X}-\mu)^2 \end{align}$$
By the Weak Law of large numbers we obtain
$$\bar{X} = \frac{1}{n} \sum_{i=1}^n X_i \to \mu \quad \text{WLN} $$
$$(\bar{X}-\mu)^2 \to 0 \quad\text{ Slutsky}$$
How do I show this term converges using only Weak Law $$ \\ \frac{1}{n} \sum_{i=1}^n (X_i-\mu)^2 \to \mathbb{E} \left( (X_i-\mu)^2 \right)=\sigma^2 \quad \text{ How?} $$