Continuity of composition of root and floor function

Can someone give me a hint to prove the continuity of the following function: $f: \mathbf{R}\to\mathbf{R}, f(x):=\sqrt{\lfloor{x^2}\rfloor}$. I already proved the continuity of the root function and that the floor function is continuous everywhere except for integers. Now my hypothesis is that that $f$ is discontinuous at integers, but I was not able to construct a $\epsilon-\delta $ argument.

The function $f$ is discontinuous at $\pm\sqrt n$, for each $n\in\Bbb N$, and at no other point. That's so because $f\left(\sqrt n\right)=n$, whereas $x\in\left(\sqrt{n-1},\sqrt n\right)\implies f(x)=n-1$. So, take $\varepsilon=1$ and then, if $\delta>0$, you take a number $x\in\left(\sqrt{n-1},\sqrt n\right)$ such that $\sqrt n-x<\delta$, and then$$\left|f(x)-f\left(\sqrt n\right)\right|=1\geqslant\varepsilon.$$