(1): I understand that $f(x) = sin(1/x)$ is continuous at every point in the open interval (0,1), but it is not uniformly continuous on this interval (e.g. take $\epsilon=2$ and set $x_n = \frac{1}{\pi/2 + 2n\pi}$ and $y_n = \frac{1}{3\pi/2 + 2n\pi}$).
(2): I also understand that a function that is continuous on a closed, bounded interval K is uniformly continuous on K, e.g. [0,1]. (Pf: suppose f is not uniformly continuous, use B-W to find two subsequences converging to same limit, etc...).
But even if $f(x) = sin(1/x)$ is continuous on a closed, bounded interval, couldn't I still set $x_n = \frac{1}{\pi/2 + 2n\pi}$ and $y_n = \frac{1}{3\pi/2 + 2n\pi}$ still exist as in (1) above, then $|x_n - y_n| \rightarrow 0$ but $|f(x_n) = f(y_n)| \geq \epsilon$? Then, f(x) would still fail to be uniformly continuous but this contradicts (2).
Could you please help me identify the gaps in my understanding? Thank you :)
I don't understand your question. You seem to be assuming that $\sin\left(\frac1x\right)$ is continuous on $[0,1]$ and therefore uniformly continuous. It is not! First of all, it is not even defined at $0$. And if you extend it to $[0,1]$, there is no way that extension will be continuous, since $\lim_{x\to0}\sin\left(\frac1x\right)$ doesn't exist.