Let $f:=\varphi$, $a:=\|f\|_2$, $b:=\|f'\|_2$, so that $\|f\|_{H^1}=a+b$. Without loss of generality, $x=0$. For all $y\in[0,1]$, we have 
$$|f(y)-f(0)|\le\int_0^y|f'(t)|dt\le\int_0^1|f'(t)|dt\le\sqrt{\int_0^1|f'(t)|^2 dt}\le b,$$
by H\"older's inequality, whence $|f(0)|\le b+|f(y)|$. So, 
\begin{equation}
	f(0)^2\le\int_0^1(b+|f(y)|)^2\le2\int_0^1(b^2+|f(y)|^2)dy\le2(b^2+a^2)\le2(a+b)^2=2\|f\|_{H^1}^2,
\end{equation}
whence 
\begin{equation}
	|f(0)|\le\sqrt2\,\|f\|_{H^1},
\end{equation}
as desired.