Let $Y_1, Y_2, ..., Y_n$ represent data from a continuous distribution $F$. The empirical distribution function $F_e$ of these data is defined as $F_e(x) = \dfrac{\sum_{i=1}^n \mathbf{1}(Y_i \leq x)}{n}$ where $\mathbf{1}(z) = 1$ if the predicate $z$ is true and 0 otherwise. Now define $D = \textrm{max}_x | F_e(x)-F(x) |$. Also define $E = \textrm{max}_{0 \leq y \leq 1} \Big|\dfrac{\sum_{i=1}^n \mathbf{1}(U_i \leq y)}{n} - y \Big|$ where $U_1, U_2, ..., U_n$ represent data from a $[0,1]$ uniform distribution. Now prove that $P(E \geq d) = P(D \geq d)$.
I have a very intuitive feel of this. I am not able to show it mathematically.
Lemma: $U=F(Y)$ is a standard uniform r.v
Let the generalized inverse distribution function: $\mathcal F^{-1}(x) = inf \{x \in \mathbb R, F(z)\geq x \}$.
$\forall u\in \mathbb R, P(U\leq u) = P(F(Y)\leq u) = P(Y\leq \mathcal F^{-1}(u)) = F(\mathcal F^{-1}(u)) = u \mathbb 1_{u\in[0,1] + \mathbb 1_{u>1} } \\$
Note: You can draw a graph to convince you that $\forall u\in[0,1], F(\mathcal F^{-1}(u)) = u$ even when F is constant on some interval.
Proof: $P(E \geq d) = P(D \geq d)$
This equation is equivalent to say that E and D have the same distribution: $ E\overset{\mathcal D}= D$
Using our lemma:
$ \forall x \in \mathcal D_Y, F_e(x) = \dfrac{\sum_{i=1}^n \mathbf{1}(Y_i \leq x)}{n} = \dfrac{\sum_{i=1}^n \mathbf{1}(F(Y_i) \leq F(x))}{n} \overset{D}= \dfrac{\sum_{i=1}^n \mathbf{1}(U_i \leq F(x))}{n}$
Note also that $\forall u\in [0,1], \exists x\in \mathbb R$ s.t $F(x) =u$.
Therefore for any function g: $\underset{u\in[0,1]}{max} (g(u)) = \underset{x\in \mathbb R}{max} (g(F(x))) $
Using our two results consecutively:
$\underset{u\in[0,1]}{max} |\dfrac{\sum_{i=1}^n \mathbf{1}(U_i \leq u)}{n} - u| = \underset{x\in \mathbb R}{max} |\dfrac{\sum_{i=1}^n \mathbf{1}(U_i \leq F(x))}{n}-F(x)| \overset{D}= \underset{x\in \mathbb R}{max} |\dfrac{\sum_{i=1}^n \mathbf{1}(Y_i \leq x)}{n}-F(x)| $