So as stated in the title I need the boundary of the set in $\mathbb{R}^2$ of $0<x<1, y=\sin(1/(1-x))$. I understand what S looks like (the curve) and therefore the curve between $0$ and $1$ is apart of the boundary (because $y$ equals the sin function). However, I get fuzzy around $x=1$. $y$ oscillates with infinite frequency between $0$ and $1$, and I believe there doesn't exist a limit. So what happens to the boundary of the set at that point?
Thanks!
P.S. - S interior is empty because it all belongs to the boundary right?
Yes, the interior is empty.
But as you note, at $x = 1~$ things are tricky. The whole set $$ \{ (1, y) : -1 \le y \le 1 \} $$ is also contained in the boundary. Here's why: for each such point $(1, y_0)$, there's a sequence of points in the set that converges to it. (Just draw a horizontal line $y = y_0$; the intersections of that line with the sine-like curve form a sequence that converges to $(1, y_0)$).
To be even more explicit: let $y_0$ be a value between $-1$ and $1$, and let $t = \arcsin(y_0)$. Then the points of the sine-like curve at $$ x = 1 - \frac{1}{t + 2 \pi k}, k = 1, 2, 3, \ldots $$ all are of the form $(x, y_0)$ and converge to $(1, y_0)$.