Consider an i.i.d. sample $X_{1}, \dots, X_{n}$ from a distribution with CDF $F\left(x\right)$, and let
\begin{equation}
\widehat{F}_{n}\left(x\right) = \dfrac{1}{n}\sum_{i = 1}^{n}\mathbb{I}_{X_{i} \leq x}
\end{equation}
denote the usual empirical distribution function. The Dvoretzky-Kiefer-Wolfowitz inequality states that
\begin{equation}
\operatorname{Pr}\left(\sup_{x\in\mathbb{R}}\left\{\sqrt{n}\left|\widehat{F}_{n}\left(x\right) - F\left(x\right)\right|\right\} > \epsilon\right) \leq 2e^{-2\epsilon^{2}}
\end{equation}
for all $\epsilon > 0$. To provide some intuition behind this result, first recognise that for any $x$ such that $F\left(x\right) \in \left(0, 1\right)$, the Central Limit Theorem (or alternatively using the weaker De Moivre–Laplace theorem) gives that
\begin{equation}
\sqrt{n}\left(\widehat{F}_{n}\left(x\right) - F\left(x\right)\right) \overset{\mathrm{d}}{\to} \mathcal{N}\left(0, F\left(x\right)\left(1 - F\left(x\right)\right)\right)
\end{equation}
since $\widehat{F}_{n}\left(x\right)$ is essentially the sample proportion of observations not greater than $x$, which we can compute to have variance
\begin{equation}
\operatorname{Var}\left(\widehat{F}_{n}\left(x\right)\right) = \dfrac{F\left(x\right)\left(1 - F\left(x\right)\right)}{n}
\end{equation}
(eg. by the binomial distribution). The takeaway here is that $\widehat{F}_{n}\left(x\right)$ will be approximately Gaussian for large $n$. Now, we use the fact that for a Gaussian random variable $Y \sim \mathcal{N}\left(\mu, \sigma^{2}\right)$, a well-known two-sided concentration inequality is given by
\begin{equation}
\operatorname{Pr}\left(\left|Y - \mu\right| > \varepsilon\right) \leq 2\exp\left(-\dfrac{\varepsilon^{2}}{2\sigma^{2}}\right)
\end{equation}
for all $\varepsilon > 0$. As an aside, this inequality can be proven using the moment generating function of a Gaussian distribution in combination with the Chernoff bound to obtain the one-sided tail inequality, and then Boole's inequality for the two-sided tail inequality. See for example (2.9) of here to obtain further details.
Here is where the 'intuitive' step comes in. If we treat $\widehat{F}_{n}\left(x\right)$ as actually being Gaussian and apply the concentration inequality for Gaussians, we will recover the Dvoretzky-Kiefer-Wolfowitz inequality. Note that $F\left(x\right)\left(1 - F\left(x\right)\right)$ is maximised at $F\left(x\right) = 1/2$, hence $\sup_{x\in\mathbb{R}}F\left(x\right)\left(1 - F\left(x\right)\right) = 1/4$. So this gives
\begin{align}
\operatorname{Var}\left(\widehat{F}_{n}\left(x\right)\right) &= \dfrac{F\left(x\right)\left(1 - F\left(x\right)\right)}{n} \\
&\leq \dfrac{1}{4n}
\end{align}
Treating $\widehat{F}_{n}\left(x\right)$ as Gaussian and plugging it into the concentration inequality gives for any $x \in \mathbb{R}$:
\begin{equation}
\operatorname{Pr}\left(\left|\widehat{F}_{n}\left(x\right) - F\left(x\right)\right| > \varepsilon\right) \leq 2\exp\left(-\dfrac{\varepsilon^{2}}{2\operatorname{Var}\left(\widehat{F}_{n}\left(x\right)\right)}\right)
\end{equation}
Taking the worst-case supremum over $x$ on either side separately:
\begin{align}
\operatorname{Pr}\left(\sup_{x\in\mathbb{R}}\left|\widehat{F}_{n}\left(x\right) - F\left(x\right)\right| > \varepsilon\right) &\leq 2\exp\left(-\dfrac{\varepsilon^{2}}{2\sup_{x'\in\mathbb{R}}\operatorname{Var}\left(\widehat{F}_{n}\left(x'\right)\right)}\right) \\
&\leq 2\exp\left(-\dfrac{\varepsilon^{2}}{2/\left(4n\right)}\right) \\
&= 2e^{-2n\varepsilon^{2}}
\end{align}
Applying the substitution $\epsilon = \sqrt{n}\varepsilon$ then yields the Dvoretzky-Kiefer-Wolfowitz inequality as originally claimed:
\begin{equation}
\operatorname{Pr}\left(\sqrt{n}\sup_{x\in\mathbb{R}}\left|\widehat{F}_{n}\left(x\right) - F\left(x\right)\right| > \epsilon\right) \leq 2e^{-2\epsilon^{2}}
\end{equation}
