Check proof that $\sin\sqrt{|x|}$ is not periodic I want to:

Prove that $\sin\sqrt{|x|}$ is not periodic

A similar question had already been asked here. But the accepted answer uses definition of the derivative. And i'm trying to do it in a "pre-calculus" manner.
Here is my try. By definition of periodic functions:
$$
f(x) = \sin\sqrt{|x|} = \sin\sqrt{|x - T|}
$$
This may be rewritten as:
$$
f(x) = \cases{\sin\sqrt{x}, \; x \ge 0 \\
\sin\sqrt{-x}, \; x < 0
}
$$
On the other hand:
$$
f(x) = \cases{\sin\sqrt{x-T}, \; x-T \ge 0 \iff x \ge T \\
\sin\sqrt{T-x}, \; x-T < 0 \iff x < -T
}
$$
So for the first case i have 
$$\sin\sqrt{x} = \sin\sqrt{x-T}$$, but $\forall{T} > 0, \exists x \ge 0 : x < T$ which contradicts the fact that $x \ge T$. The second case is handled similarly.
Is it valid?

It is not correct. There is no reason to assume that when we are in the first case in the case of $\sin\left(\sqrt{|x|}\right)$, then we are also in the first case in the case of $\sin\left(\sqrt{|x-T|}\right)$ and vice-versa.
It's quite easy to deduce that the function is not periodic from the fact that its zeros form the set$$\left\{n^2\pi^2\,\middle|\,n\in\mathbb{Z}^+\right\}.$$
