$\sqrt{n}\sup_x|F_n-F|=\sup_x|\frac{1}{\sqrt{n}}\sum_{i=1}^nZ_i(x)| $
where $Z_i(x)=1_{X_i\leq x}-E[1_{X_i\leq x}]$
by CLT you have
$G_n=\frac{1}{\sqrt{n}}\sum_{i=1}^nZ_i(x)\rightarrow \mathcal{N}(0,F(x)(1-F(x)))$
this is the intuition...
brownian bridge $B(t)$ has variance $t(1-t)$ http://en.wikipedia.org/wiki/Brownian_bridge replace $t$ by $F(x)$. This is for one $x$...
You also need to check the covariance and hence it still is easy to show (CLT) that
for ($x_1,\dots,x_k$)
$(G_n(x_1),\dots,G_n(x_k))\rightarrow (B_1,\dots,B_k)$ where $(B_1,\dots,B_k)$ is $\mathcal{N}(0,\Sigma)$ with $\Sigma=(\sigma_{ij})$, $\sigma_{ij}=\min(F(x_i),F(x_j))-F(x_i)F(x_j)$.
The difficult part is to show that the distribution of the suppremum of the limit is the supremum of the distribution of the limit... Understanding why this happens requires some empirical process theory, reading books such as van der Waart and Welner (not easy). The name of the Theorem is Donsker Theorem http://en.wikipedia.org/wiki/Donsker%27s_theorem ...