By the Dede Moivre–Laplace theorem, \begin{equation} P(S_n=k)=P(B_n=(n+k)/2)\sim\frac1{\sqrt{\pi n/2}}\,\exp\{-k^2/(2n)\}, \end{equation} where $B_n$ is a random variable with the binomial distribution with parameters $n$ and $1/2$ and $k=O(\sqrt n)$. Here and in what follows, $k$ and $m$ are integers which equal $n\text{ mod }2$, and $n\to\infty$.
So, by the reflection principle (Theorem 0.8, page 4), for
\begin{equation}
M_n:=\max_{0\le j\le n}S_j
\end{equation}
we have
\begin{align*}
P(M_n\ge m,S_n=k)=P(S_n=2m-k)&\sim P(S_n=k)\exp\{-\tfrac1{2n}((2m-k)^2-k^2)\} \\
&\sim P(S_n=k)\exp\{-2z(z-u)\}
\end{align*}
if $m\sim z\sqrt n$, $k=(u+o(1))\sqrt n$, $z$ and $u$ are real numbers, and $z>0\vee u$, so that
\begin{align*}
P(M_n\ge m|S_n=k)\to\exp\{-2z(z-u)\},
\end{align*}
and obviously $\exp\{-2z(z-u)\}\to0$ asif $z\to\infty$ and $u=O(1)$.
Therefore,
\begin{align*}
P(M_n\ge m|S_n=k)\to0
\end{align*}
uniformly in $k$ if $k=O(\sqrt n)$ and $m/\sqrt n\to\infty$.
So, letting
\begin{equation}
M^*_n:=\max_{0\le j\le n}|S_j|,
\end{equation}
by the symmetry we have
\begin{align*}
P(M^*_n\ge m|S_n=k)\le P(M_n\ge m|S_n=k)+P(M_n\ge m|S_n=-k)\to0
\end{align*}
and hence
\begin{align*}
P(M^*_n\ge m|\,|S_n|\le|k|)=\sum_{j\colon|j|\le |k|}P(M^*_n\ge m|S_n=j)P(S_n=j)\to0,
\end{align*}\begin{align*}
P(M^*_n\ge m|\,|S_n|\le|k|)=\sum_{j\colon|j|\le |k|}P(M^*_n\ge m|S_n=j)P(S_n=j|\,|S_n|\le|k|)\to0,
\end{align*}
if $k=O(\sqrt n)$ and $m/\sqrt n\to\infty$; that is, $M^*_n=O_P(\sqrt n)$ conditionally on $S_n=O(\sqrt n)$.
In particular, it follows that for even $n$ we have $S_{n/2}=O_P(\sqrt n)$ conditionally on $S_n=O(\sqrt n)$, as desired.