Function $\mathbb{R}\to\mathbb{R}$ that is continuous and bounded, but not uniformly continuous

To prove that $f(x)=\sin(x^2)$ is not uniformly continuous, let $\epsilon=\frac{1}{2}$. You want to show that for every $\delta\gt 0$ there exists $x$ and $y$ (which may depend on $\delta$) such that $|x-y|\lt \delta$, but $|f(x)-f(y)|\geq \frac{1}{2}$. Suggestion: try to pick a $y$ where $f(y)=0$, and a very nearby $x$ where $f(1)=\pm 1$. You want $y^2=n\pi$ for some integer $n$, and $x^2$ to be $\frac{1}{2}\pi$ away. If you cannot produce a pair explicitly, maybe you can show that you can pick them as close as you want anyway, provided $x$ and $y$ are large enough (which agrees with your observation that the derivative is unbounded).

Let $T_n(x) = 1-n|x|$ for x in $[-1/n, 1/n]$. As $T_n$ is linear, the required $\delta$ is $\epsilon/n$. Try to construct a continuous function by using combinations of these $T_n$'s and constant (zero) functions. I'm imagining a graph where every once in a while there is a spike with graph $T_n$, but as you go out to infinity, the spikes get steeper.