If $x = 0$ or $y = 0$, then the claim clearly holds with equality, so assume that $x \neq 0 \neq y$. Then $2 \sqrt{x y} > 0$, so that
\begin{equation}
x + y
< x + 2 \sqrt{x y} + y
\end{equation}
Applying result
R4261: Real binomial theorem for exponent two to the expression $x + 2 \sqrt{x y} + y$ with $a : = \sqrt{x}$ and $b : = \sqrt{y}$, we have
\begin{equation}
x + 2 \sqrt{x y} + y
= (\sqrt{x} + \sqrt{y})^2
\end{equation}
Thus, applying
R4260: Real square root function is strictly isotone, we conclude
\begin{equation}
\sqrt{x + y}
< \sqrt{x + 2 \sqrt{x y} + y}
= \sqrt{(\sqrt{x} + \sqrt{y})^2}
= (\sqrt{x} + \sqrt{y})^{\frac{1}{2} \cdot 2}
= \sqrt{x} + \sqrt{y}
\end{equation}
$\square$
