Before this is marked as a duplicate, I have already looked at the answers here and here.
I believe my question is different because I want to do this proof directly from the definition. The proof given at the first link uses the property of compactness, Lipschitz continuity and derivatives to support the argument. However, my professor's real analysis approach has not been that of metric spaces and, additionally, Lipschitz continuity has not been discussed yet so I'd like to avoid them for now and attempt this proof directly from the definition. I'm interested in a proof similar to the second answer given on the first link but, unfortunately, there are no comments on whether or not it is correct.
The question is to determine whether or not $f(x) = x \sin{\frac{1}{x}}$ is uniform continuous on $(1,2)$.
I present my own attempt:
Given $\epsilon>0$, let $\delta= \frac{\epsilon-1}{2}$ Then $|x-y|<\delta \implies$ $$\left | f(x)-f(y)\right | =\left | x\sin \left(\frac{1}{x}\right)-y\sin \left(\frac{1}{y}\right) \right|= \left |(x-y) \left( \sin \left(\frac{1}{x}\right)+\sin \left(\frac{1}{y}\right) \right) -x\sin \left(\frac{1}{y} \right)+y\sin \left(\frac{1}{x} \right)\right| \leq \left |(x-y) \left( \sin \left(\frac{1}{x}\right)+\sin \left(\frac{1}{y}\right) \right)\right| + \left|-x\sin \left(\frac{1}{y} \right)+y\sin \left(\frac{1}{x} \right)\right|$$
$$\leq \left |x-y \right| \left |\left( \sin \left(\frac{1}{x}\right)+\sin \left(\frac{1}{y}\right) \right)\right| + \left|-x\sin \left(\frac{1}{y} \right)+y\sin \left(\frac{1}{x} \right)\right|$$
$$\leq 2\left |x-y \right|+1 < 2\cdot (\frac{\epsilon-1}{2})+1=\epsilon$$
Any help is appreciated.
The problem with your approach is that $\delta = \frac{\epsilon - 1}{2}$ is not a very good choice of $\delta$: what if $\epsilon-1<0$? Here is a correct solution.
We have that $|\sin(s) - \sin(t)| = \left|2\overbrace{\sin\left(\frac{s-t}{2}\right)}^{\leq \frac{s-t}{2}}\underbrace{\cos\left(\frac{s+t}{2}\right)}_{\leq 1}\right| \leq |s-t|$ for any real $s,t$.
This implies that $|\sin \frac{1}{x} - \sin \frac{1}{y}| \leq \left|\frac{1}{x} - \frac{1}{y}\right| = \left|\frac{x-y}{xy}\right| \leq |x-y|$ since $xy \geq 1$, since $x,y \in (1,2)$.
Consequently, observe that $$\left|x\sin \frac{1}{x} - y \sin \frac{1}{y}\right| = \left|x\left(\sin \left(\frac{1}{x}\right) - \sin\left(\frac{1}{y}\right)\right) + \sin\left(\frac{1}{y}\right)(x-y)\right| \leq \overbrace{|x|}^{\leq 2}\underbrace{\left|\sin \left(\frac{1}{x}\right) - \sin\left(\frac{1}{y}\right)\right|}_{\leq |x-y|}$$ $$+ \ \underbrace{\left|\sin\left(\frac{1}{y}\right)\right|}_{\leq 1}|x-y| \leq 3|x-y| < \epsilon$$ provided we choose $\delta = \frac{\epsilon}{3}$.