Showing the empirical cdf to be consistent

Say we have the empirical c.d.f. of $X_{i}$ to be $\hat{F}_{n}\left(x\right)=\frac{1}{n}\sum_{1\leq i\leq n}I\left\{ X_{i}\leq x\right\}$. How can we show that this converges in probability to the true c.d.f. that is to show $\forall\varepsilon>0,\quad\lim_{n\to\infty}P\left\{ \left|\hat{F}_{n}\left(x\right)-F\left(x\right)\right|>\varepsilon\right\} =0$? Asymptotic Statistics by van der Vaart says to use the Strong Law of Large Numbers but I don't really get how that works. Is there a simpler way of showing it with the Weak Law of Large Numbers. I've tried showing it with Markov's Inequality as well, but I've gotten stuck there.

Let $Y_i = I\{X_i \leq x\}$. Then, the $Y_i$ are i.i.d. with mean $F(x)$. By the strong law of large numbers (you'll have to check some other condition on $Y_i$ beyond finite mean -- see Durret's Probability book or any other intro probability book, but it suffices to note that $E[Y_i^4] \leq 1$, so you can apply the strong law of large numbers), $\hat{F}_n(x) = \frac{Y_1 + Y_2 + \ldots+ Y_n} {n} \to E[Y_i] = F(x)$ almost surely. Note convergence in the almost sure sense implies convergence in probability, so you have $\hat{F}_n(x) \to F(x)$ in probability as well.
If you switch the law of large numbers, you'll get different versions of convergence, but the idea is basically the same: $\hat{F}_n(x) = \frac{Y_1 + Y_2 + \ldots+ Y_n} {n}$ is an average of i.i.d. random variables, which converges in some appropriate sense (depending on which law of large numbers you apply) to $E[Y_1] = F(x)$. You could apply a weak law of large numbers and get convergence in probability directly.