Possible error in Lipschutz The problem asks to show that $f:(0,\infty)\to [-1,1]$ given by $f(x)=\sin (\frac {1}{x})$ is neither open nor closed. Where $(0,\infty)$ and $[-1,1]$ are subspaces of $\Bbb{R}$ with the usual topology. 
While it is clear this is true if $[-1,1]$ is replaced with $\Bbb R$ , the images of open set will look like $[-1,a)$ , $(a,1]$ , or $[-1,-1]$ all of which are open in $[-1,1]$. 
However the map is not closed since the image of $[\frac 1 {2\pi} , \infty)$ is $(0,1]$ which is not closed in $[-1,1]$.
So why is this map not open?

It is indeed an error. Probably the author thought of "open in $\mathbb{R}$", even though explicitly mentioning the subspace topology of $[-1,1]$.
Since $x\mapsto \frac{1}{x}$ is a homeomorphism $(0,\infty) \to (0,\infty)$, $f$ is open [or closed] if and only if $g(x) = \sin x$ is open [resp. closed] as a map $(0,\infty) \to [-1,1]$.
It is relatively easy to show that $\sin (I)$ is open in $[-1,1]$ for every open interval $I \subset (0,\infty)$. For points with $-1 < \sin x < 1$, the function is a homeomorphism on a small enough neighbourhood of $x$ [so small that it contains no point with $\lvert \sin x\rvert = 1$], and if $\sin x = \pm 1$, for $0 < \varepsilon < \pi$, we have $\sin (x-\varepsilon,x+\varepsilon) = (\cos \varepsilon, 1]$ resp. $\sin (x-\varepsilon,x+\varepsilon) = [-1,-\cos \varepsilon)$, which are open in $[-1,1]$.
