By the [de Moivre–Laplace theorem][1], 
\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)][2], 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 $\exp\{-2z(z-u)\}\to0$ if $z\to\infty$ 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|\,|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. 

[1]: https://en.wikipedia.org/wiki/De_Moivre%E2%80%93Laplace_theorem

[2]: https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=1&ved=2ahUKEwj97vXNl8veAhUJOawKHTMmAYIQFjAAegQICRAC&url=http%3A%2F%2Fcgm.cs.mcgill.ca%2F~breed%2FMATH671%2Flecture2corrected.pdf&usg=AOvVaw2AWnEgYglDpWEJX383zQ14