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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 $S$.

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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...

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    This is what I've done too. But my concern is that if $k \to \infty$ then $x \to -\infty$ But $\lim_{x \to -\infty} \sin e^{-x} \neq 0$. So I got confused. – 2017-01-27
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    @VikrantDesai Why do you care about that limit $\;\lim\limits_{x\to-\infty}e^{-x}\;$ ?? That is irrelevant to the problem. What is relevant is that there are infinite points of the form $\;x=-\log\pi k \;$ for which $\;\sin e^{-x}=0\;$ and *these points' limit* is $\;-\infty\;$ , so that then the set $\;\{(-\log\pi k\,,\,0)\}\subset\Bbb R^2\;$ is unbounded. – 2017-01-27
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    Fair enough. I stop my overthinking. :D – 2017-01-27