It is instructive to see an example where Hartogs' theorem fails in one dimension. Take $U = {\mathbb C}$ and let $K = \{0\}.$ The function $\frac{1}{z}$ is holomorphic in $U \setminus K,$ but cannot be extended to $U.$

To understand the example and failure of the theorem it is important to understand the proof. In the proof, the way we construct an extension is that we start with a function holomorphic in $U \setminus K,$ modify it in a neighbourhood of $K$ to be zero, hence extending as a smooth function through $K.$ Then we solve the $\bar{\partial}$ operator inhomogeneous equation $\bar{\partial}\psi = g$ to ``correct'' our extension to be holomorphic. The key point is that $g$ has compact support allowing us to solve the equation and find a $\psi$ with compact support. This fails in dimension 1. While we always get a solution $\psi,$ the solution can never have compact support. Hence, if we tried the proof with $\frac{1}{z},$ the new function we obtain in the proof does not agree with $\frac{1}{z}$ on any open set and hence is not an extension.