Let \(k = \left[ \sqrt{[x]} \right]\):
$$\begin{gather} k \le \sqrt{[x]} \lt k+1 \\ k^2 \le [x] \lt (k+1)^2 \end{gather}$$
Since \([x]\) is an integer:
$$\begin{gather} k^2 = [x] \\ k = \sqrt{[x]} \end{gather}$$
Since \(k = \left[ \sqrt{[x]} \right]\), then \(\left[ \sqrt{[x]} \right] = \sqrt{[x]}\).