Given the continuous periodic function
$$f:\mathbb{R} \to \mathbb{R}$$
with a periodic factor of $2$, so that
$$f(x+2)=f(x)$$
and
$$f(0) > f(1)$$
Show that the function $g:\mathbb{R} \to \mathbb{R}$, which is defines as $g(x) = f(x^2)$, is not uniformly continuous.
I tried writing down the definition of the uniform continuity and starting from there, I also tried assuming that $g$ is uniformly continuous to get a contradiction but I got stuck every time.
Some help would be appreciated.
Hint: For every $\delta>0$ there exists $k\in \mathbb{N}$ sufficiently large such that $$\vert \sqrt{2k} - \sqrt{2k+1}\vert = \vert \frac{(\sqrt{2k} - \sqrt{2k+1})(\sqrt{2k} + \sqrt{2k+1})}{(\sqrt{2k} + \sqrt{2k+1})}\vert = \frac{1}{\sqrt{2k}+\sqrt{2k+1}} < \frac{1}{\sqrt{2k}}<\delta.$$ Furthermore, $$ g(\sqrt{2k+1})-g(\sqrt{2k})=f(1)-f(0).$$
The definition of uniform continuity would be $$ \forall \varepsilon >0 \exists \delta>0 \forall x,y\in \mathbb{R}: \vert x -y\vert < \delta \Rightarrow \vert g(x) - g(y) \vert < \varepsilon. $$ Now we logically negate this statement to get NOT uniformly continuous: $$ \exists \varepsilon >0 \forall \delta>0 \exists x,y\in \mathbb{R} : \vert x-y\vert < \delta \wedge \vert g(x) - g(y) \vert \geq \varepsilon.$$ Now, for $\varepsilon = f(0) - f(1)$ and a given $\delta>0$ what would you need to pick such that we can show that $g$ is not uniformly continuous? Indeed, exactly $x=\sqrt{2k}$ and $y=\sqrt{2k+1}$ would work for sufficiently large $k$.