We will prove the result separately for the intervals $(0,1]$ and $\left(1,\frac{\pi}{2}\right]$.
Assume first that $1<x<\frac{\pi}{2}$. Since $0<\sin x < 1$, we have $\sqrt{\sin x}>\sin x$. Since $x>1$, we have $x>\sqrt{x}$, and, as $\sin x$ is an increasing function, $\sin x > \sin \sqrt{x}$. We may therefore conclude:
$$\displaystyle
\sqrt{\sin x}>\sin x>\sin \sqrt{x}
$$
in this case.
Assume now that $0<x\leq 1$. This implies that $x<\sqrt{x}$.
Since $\sqrt{x}>0$, the inequality is equivalent to:
$$\displaystyle
\begin{align*}
\frac{\sqrt{\sin x}}{\sqrt{x}} &> \frac{\sin \sqrt{x}}{\sqrt{x}}\\
\sqrt{\frac{\sin x}{x}} &> \frac{\sin \sqrt{x}}{\sqrt{x}} \\
\sqrt{f(x)} &> f(\sqrt{x})
\end{align*}
$$
where $f(x) = \frac{\sin x}{x}$.
Because $0<f(x)<1$ in the interval under consideration, $\sqrt{f(x)} > f(x)$.
Now, the derivative of $f(x)$ is:
$$\displaystyle
\frac{df}{dx} = \frac{x\cdot\cos x - \sin x}{x^2}
$$
and this is negative, since $0< x< \tan x$. This means that $f$ is strictly decreasing; as $x<\sqrt{x}$, $f(x) > f(\sqrt{x})$, and we conclude:
$$\displaystyle
\sqrt{f(x)} > f(x) > f(\sqrt{x})
$$
which completes the proof.