Let us name this set $S=\{(x,y)\in \Bbb R^2 : y=0, \sin(e^{-x})=0\}.$
Now, $\sin (e^{-x})=0 \Rightarrow e^{-x}=n\pi \; \forall n \in \Bbb N.$(As graph of $e^{-x}$ lies in only first and second quadrant) $\Rightarrow -x=\ln n\pi \Rightarrow x=\ln \frac {1}{n\pi}.$
Hence $S = \{( \ln \frac {1}{n\pi},0):n \in \Bbb N\}$.
Also I noticed that $\lim_{x \to \infty} \sin (e^{-x})=0.$ But don't know how to incorporate it into the solution.
Also as $n \to \infty$,we see $\ln \frac {1}{n\pi} \to -\infty.$ But this doesn't satisfy the condition $\sin (e^{-x})=0.$
Based on these observations I am confused a little bit on how to move forward. Although I know that I have to check only closedness and boundedness of $S$.
Observe that
$$\sin e^{-x}=0\iff e^{-x}=\pi k\;,\;\;k\in\Bbb N\cup\{0\}\iff x=-\log\pi k$$
and from here that $\;S\;$ isn't bounded...