Let $f:\mathbb{R}\to (0,1)$, $f=t\sin\left(\frac1t\right)$. I think this function is uniformly continuous, but I don't see how to prove this formally. I can observe that if $s,t\in (0,1)$ and $|s-t|<\delta$, then the smaller $\delta$ we take, the smaller the quantity $$\left|t\sin\left(\frac1t\right)-s\sin\left(\frac1s\right)\right|\mbox{ (*)}$$ becomes. So we can take any $\varepsilon>0$ and any $s,t$, such that (*) is less than $\varepsilon$, and there will necessarily be a $\delta>0$ such that $|s-t|<\delta$.
I would appreciate some suggestions on what approach to take.
If a function is differentiable, then it must necessarily be continuous.
Now,
$$\frac{df}{dt} = -\frac{1}{t}\cos{\frac{1}{t}} + \cos{\frac{1}{t}}$$
As you can see, the derivative is defined at all points except at $t=0$.
So for $t=0$, we will have to check the limit.
The limit is easy to calculate (by applying the squeeze theorem) and comes out to be
$$t\sin{\frac{1}{t}} = 0$$
when $t$ tends to $0$.
Hence proved.