Set $Z_n = \frac{1}{n}\sum_{i=1}^n (X_i - \mathbb{E}[X])$, so $\bar{X}_n - \mathbb{E}[X] = Z_n$. By independence of the $X_i$, the variance satisfies $\operatorname{Var}(Z_n) = \operatorname{Var}(X)/n$. Chebyshev's inequality then gives $\mathbb{P}(|\bar{X}_n - \mathbb{E}[X]| > \varepsilon) \leq \operatorname{Var}(X)/(n\varepsilon^2) \to 0$ for every fixed $\varepsilon > 0$.