Let $d$ be the usual Euclidean metric on $\Bbb R$ and $f\colon (\Bbb R, d) \to ([-1,1],d)$ be the map given by $f(0)=0$ and $f(x)=\sin(\frac1x)$ if $x\neq0$.
Let $A$ be an open subset of $\Bbb R$ such that $0 \in A$. Show that $f[A] = [-1,1]$.
My idea is to construct a proof which shows that this mapping sends open sets into open sets but $\sin(\frac1x)$ is not continuous at $0 $. My question is, how can i show that $f[A] = [-1,1]$ since there is a discontinuity at $0$? Or does this not matter in this case?
Hint.
What are the values of $f$ at points $\frac{1}{2 k\pi+\frac{\pi}{2}}$ and $\frac{1}{2 k\pi-\frac{\pi}{2}}$ for $k \in \mathbb Z$?