Not all singularities must be isolated. Consider the following examples:

1. $f(z)=\frac{1}{\sin \left(\frac{1}{z}\right)}$. We know this is undefined if $\sin \left(\frac{1}{z}\right)=0$, which happens when $\frac{1}{z}=2\pi n$, so $z=\frac{1}{2\pi n}$ or $z=0$. In this scenario, where $U=\mathbb{C}-\{\frac{1}{2\pi n}|n\in \mathbb{N}\}$, 1;;$z=0$ is not an isolated singularity.
2. There is also a family of holomorphic functions called Lacunary Power Series. One example is $f(z)=\sum _{i=0}^{\mathrm{\infty }}{z}^{2^n}$. By the comparison test, this series converges for $|z|<1$. Now we also note that for $z$ converging up to $1$ on the real axis, we have $f(z)$ diverging to $\mathrm{\infty }$. Similarly, if $z$ converges down to $-1$, $f(z)$ diverges to $\mathrm{\infty }$. Now looking at $f(z)=z+(1+z^2+z^4+\dots )=z+f(z^2)$, we notice that the end behavior just depends on $f(z^2)$. So as $z$ converges in a straight line to $\pm i$, we have that this diverges. Finally, we extend this process to show that $f(z)$ diverges at any of the points of the form
3. $$e^{\frac{(i\pi m)}{2^n}}\$\$Thedyadicintegersaredenseontheunitinterval,sotheyaredenseontheunitcirclebythenatureofthishomeomorphism.Eachoneofthesepointsisanon-isolatedsingularity,andthecirclewhichthesepointsboundiscalleda"naturalboundary"of\$f\$.$$