[proofplan]
Fix $x\in\mathbb R$ and rewrite the empirical count $nF_n(x,\cdot)$ as a sum of indicator random variables recording whether each observation falls in $(-\infty,x]$. These indicators are independent Bernoulli random variables with success probability $F(x)$. We then compute the probability that their sum equals $k$ by choosing which $k$ of the $n$ indicators equal $1$, obtaining the binomial probability mass function.
[/proofplan]
[step:Convert the empirical count into a sum of Bernoulli indicators]
Fix $x\in\mathbb R$ and define $p:=F(x)$. The symbols $(\Omega,\mathcal A,\mathbb P)$ denote the probability space from the theorem statement, and $\omega\in\Omega$ denotes a sample point. For each $i\in\{1,\dots,n\}$, the event $\{\omega\in\Omega:X_i(\omega)\le x\}$ belongs to $\mathcal A$ because $X_i:(\Omega,\mathcal A)\to(\mathbb R,\mathcal B(\mathbb R))$ is measurable and $(-\infty,x]$ is a Borel subset of $\mathbb R$. Define the [random variable](/page/Random%20Variable) $Y_i:\Omega\to \{0,1\}$ by
\begin{align*}
Y_i(\omega)=\mathbb{1}_{\{X_i\le x\}}(\omega).
\end{align*}
Then
\begin{align*}
\mathbb P(Y_i=1)=\mathbb P(X_i\le x)=F(x)=p,
\end{align*}
and
\begin{align*}
\mathbb P(Y_i=0)=1-p.
\end{align*}
Thus each $Y_i$ has Bernoulli distribution with parameter $p$. Moreover, since $X_1,\dots,X_n$ are independent and each $Y_i$ is a measurable function of $X_i$, the random variables $Y_1,\dots,Y_n$ are independent.
By the definition of the empirical distribution function $F_n$,
\begin{align*}
nF_n(x,\omega)=\sum_{i=1}^n Y_i(\omega)
\end{align*}
for every $\omega\in\Omega$.
[/step]
[step:Compute the distribution of the Bernoulli sum]
Let $S:\Omega\to \{0,1,\dots,n\}$ be the random variable defined by
\begin{align*}
S(\omega)=\sum_{i=1}^n Y_i(\omega).
\end{align*}
For $k\in\{0,1,\dots,n\}$ and a subset $A\subset\{1,\dots,n\}$ with $|A|=k$, define the event
\begin{align*}
E_A:=\{\omega\in\Omega:Y_i(\omega)=1\text{ for }i\in A,\;Y_j(\omega)=0\text{ for }j\notin A\}.
\end{align*}
By independence of $Y_1,\dots,Y_n$,
\begin{align*}
\mathbb P(E_A)=\prod_{i\in A}\mathbb P(Y_i=1)\prod_{j\notin A}\mathbb P(Y_j=0).
\end{align*}
Substituting $\mathbb P(Y_i=1)=p$ for $i\in A$ and $\mathbb P(Y_j=0)=1-p$ for $j\notin A$ gives
\begin{align*}
\mathbb P(E_A)=p^k(1-p)^{n-k}.
\end{align*}
The events $E_A$ with $|A|=k$ are pairwise disjoint, and their union is exactly $\{S=k\}$. Since there are $\binom nk$ subsets of $\{1,\dots,n\}$ of cardinality $k$, finite additivity gives
\begin{align*}
\mathbb P(S=k)=\binom nk p^k(1-p)^{n-k}.
\end{align*}
[guided]
We want the law of the random variable $S=\sum_{i=1}^n Y_i$. To know that law, it is enough to compute $\mathbb P(S=k)$ for each possible value $k\in\{0,1,\dots,n\}$.
Fix such a $k$. The event $\{S=k\}$ means exactly that precisely $k$ of the indicators equal $1$ and the remaining $n-k$ indicators equal $0$. To record which indicators equal $1$, let $A\subset\{1,\dots,n\}$ be a subset with $|A|=k$, and define
\begin{align*}
E_A:=\{\omega\in\Omega:Y_i(\omega)=1\text{ for }i\in A,\;Y_j(\omega)=0\text{ for }j\notin A\}.
\end{align*}
On $E_A$, the successful indices are exactly the elements of $A$.
Because $Y_1,\dots,Y_n$ are independent and each $Y_i$ has Bernoulli parameter $p$, the probability of this particular pattern is
\begin{align*}
\mathbb P(E_A)=\prod_{i\in A}\mathbb P(Y_i=1)\prod_{j\notin A}\mathbb P(Y_j=0).
\end{align*}
Using $\mathbb P(Y_i=1)=p$ for every $i\in A$ and $\mathbb P(Y_j=0)=1-p$ for every $j\notin A$, this becomes
\begin{align*}
\mathbb P(E_A)=p^k(1-p)^{n-k}.
\end{align*}
Different subsets $A$ give disjoint patterns: one outcome cannot have two different sets of successful indices. Also, every outcome with $S=k$ has exactly one successful-index set $A$. Hence $\{S=k\}$ is the disjoint union of the events $E_A$ over all subsets $A\subset\{1,\dots,n\}$ with $|A|=k$. There are $\binom nk$ such subsets $A$, so finite additivity gives
\begin{align*}
\mathbb P(S=k)=\binom nk p^k(1-p)^{n-k}.
\end{align*}
This is exactly the binomial probability mass function with parameters $n$ and $p$.
[/guided]
[/step]
[step:Identify the empirical count with the binomial random variable]
From the first step, $nF_n(x,\cdot)=S$ as random variables on $\Omega$. From the second step, for every $k\in\{0,1,\dots,n\}$,
\begin{align*}
\mathbb P(S=k)=\binom nk p^k(1-p)^{n-k}.
\end{align*}
Since $p=F(x)$, this gives
\begin{align*}
\mathbb P(nF_n(x,\cdot)=k)=\binom nk F(x)^k(1-F(x))^{n-k}
\end{align*}
for every $k\in\{0,1,\dots,n\}$. Therefore $nF_n(x,\cdot)$ has binomial distribution with parameters $n$ and $F(x)$, which we write as
\begin{align*}
nF_n(x,\cdot)\sim \operatorname{Bin}(n,F(x)).
\end{align*}
[/step]
