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?

You are working at $x\ge T$, so, how can you suppose that exist some $x<T$? It doesn't make sense.
A suggestion could be use the identity:
$$\sin p-\sin q=2\sin\left(\frac{p-q}{2}\right)\cos\left(\frac{p+q}{2}\right).$$
In your case you have $p=\sqrt{x}$ and $q=\sqrt{x-T}$, for $x\ge T.$ So,
$$\frac{\sqrt{x}-\sqrt{x-T}}{2}=k\pi,\quad k\in \Bbb Z$$ or
$$\frac{\sqrt{x}+\sqrt{x-T}}{2}=\frac \pi 2+k\pi,\quad k\in \Bbb Z$$
Can you finish?
