We assume that \(\varepsilon\leq 1-\pi(H)\text{;}\) otherwise, the theorem holds vacuously, as no such graph can exist. By definition of \(\pi(H)\text{,}\) we may choose \(n_0\) large enough so that \(\ex(n,H)\leq \left(\pi(H)+\frac{\varepsilon}{2}\right)\binom{n}{2}\) for all \(n\geq n_0\text{.}\) Now, for \(n\geq n_0\text{,}\) let \(G\) be a graph with \(n\) vertices which satisfies the hypotheses of the theorem.
By “double counting,” we have that
\begin{equation*}
\binom{n-2}{n_0-2}|E(G)| = \sum_{\substack{S\subseteq V(G)\\ |S|=n_0} }e(S)\text{.}
\end{equation*}
This is because every edge of \(G\) is contained in precisely \(\binom{n-2}{n_0-2}\) sets \(S\subseteq V(G)\) of cardinality \(n_0\text{.}\) Now, say that a subset \(S\subseteq V(G)\) of cardinality \(n_0\) is dense if \(G[S]\) contains more than \(\left(\pi(H)+\frac{\varepsilon}{2}\right)\binom{n_0}{2}\) edges and sparse otherwise. Let \(\mathcal{D}\) and \(\mathcal{S}\) be the collection of all dense or sparse sets, respectively. We have
\begin{equation*}
\binom{n-2}{n_0-2}|E(G)| = \sum_{S\in\mathcal{D}}e(S) + \sum_{S\in\mathcal{S}}e(S)
\end{equation*}
\begin{equation*}
\leq |\mathcal{D}|\binom{n_0}{2} + |\mathcal{S}|\left(\pi(H)+\frac{\varepsilon}{2}\right)\binom{n_0}{2}\text{.}
\end{equation*}
If we write \(\alpha:=|\mathcal{D}|/\binom{n}{n_0}\) and apply the hypothesis on \(G\text{,}\) then
\begin{equation*}
\left(\pi(H)+\varepsilon\right)\binom{n-2}{n_0-2}\binom{n}{2} \leq \alpha \binom{n}{n_0}\binom{n_0}{2} + (1-\alpha)\left(\pi(H)+\frac{\varepsilon}{2}\right)\binom{n}{n_0}\binom{n_0}{2}\text{.}
\end{equation*}
Now, \(\binom{n-2}{n_0-2}\binom{n}{2}=\binom{n}{n_0}\binom{n_0}{2}\) (this can be easily shown algebraically, but there is also a simple “combinatorial proof”). Thus, solving for \(\alpha\) gives
\begin{equation*}
\alpha\geq \frac{\varepsilon}{2\left(1-\pi(H)-\frac{\varepsilon}{2}\right)}\text{.}
\end{equation*}
So,
\begin{equation*}
|\mathcal{D}|\geq \frac{\varepsilon}{2\left(1-\pi(H)-\frac{\varepsilon}{2}\right)}\binom{n}{n_0} \geq c' n^{n_0}
\end{equation*}
where \(c'\) is a constant depending on \(\varepsilon\) and \(H\) only (note that, in some sense, \(c'\) also depends on \(n_0\text{,}\) but that \(n_0\) itself is a function of \(\varepsilon\) and \(H\text{;}\) thus, \(c'\) can be thought of as depending on \(\varepsilon\) and \(H\) only).
Now, by definition of \(n_0\) and \(\mathcal{D}\text{,}\) for every \(S\in \mathcal{D}\text{,}\) the set \(G[S]\) must contain a copy of \(H\text{.}\) Also, any copy of \(H\) is contained in at most \(\binom{n-|V(H)|}{n_0-|V(H)|}\) sets of cardinality \(n_0\text{.}\) Therefore, the number of copies of \(H\) in \(G\) is at least
\begin{equation*}
|\mathcal{D}|/\binom{n-|V(H)|}{n_0-|V(H)|}\geq c n^{|V(H)|}
\end{equation*}
for some constant \(c\) which depends on \(\varepsilon\) and \(H\) only.
