Given the function $f(z)=\sin\left(\frac{1}{\sin(\frac{1}{z})}\right)$ it is easy to see that $z=0$ is a non isolated singularity and that $z=\frac{1}{k\pi}$ with $k\in\mathbb{Z}\setminus\{0\}$ are isolated singularities in $D'\left(\frac{1}{k\pi},r\right)$ with $r=\left|\frac{1}{(k+1)k\pi}\right|$.
The thing is that I have the next argument in my notes and I am not sure of what is being done here and if it can be done without the open map theorem.
Let $V_m$ be a neigbourhood of $\frac{1}{m\pi}$, so $U_m\{u:\frac{1}{u}\in V_m\}$ is a neighbourhood of $m\pi$.
By the open map theorem, applied to the function $sin:\mathbb{C}\to \mathbb{C}$ we have that $W_m:=\sin(U_m)$ is a neigbourhood of $sin(m\pi)=0$.
Let $g(z)=\sin(\frac{1}{z})$, we know that $g$ has an essential singularity at $z=0$, so by the Casorati-Weierstrass theorem we have that $g(W_m\setminus\{0\})$ is a dense subset of $\mathbb{C}$.
That means that for every $\alpha\in\mathbb{C}$, there is a sequence $\{w_k\}_{k\in\mathbb{N}}\in W_m\setminus\{0\}$ such that $g(w_k)$ tends to $\alpha$.
By the definition of $W_m$ we have that there is a sequence $\{u_k\}_{k\in\mathbb{N}}\in U_m\setminus\{m\pi\}$ such that $\sin(u_k)=w_k\neq 0$. By the definition of $U_m$ we have that there is a sequence $\{v_k\}_{k\in\mathbb{N}}\in V\setminus \{\frac{1}{m\pi}\}$ such that $u_k=\frac{1}{v_k}$.
Then we have that as $k\to\infty$ $$f(v_k)=\sin\left(\frac{1}{\sin(\frac{1}{v_k})}\right)=\sin\left(\frac{1}{\sin(u_k)}\right)=\sin\left(\frac{1}{w_k}\right)\to\alpha$$
So to my understanding this means that $z=\frac{1}{m\pi}$ is a essential singularity.
I am not sure if the above argument is correct and in that case as I mentioned above, there is a way to do this without the open map theorem? Why can't I start with a neighbourhood of $\sin(m\pi)=0$ and as the sine function is continuous I have neighbourhood of $m\pi$ and argue this way without the open map theorem?