Let $\Omega$ be an open set in $\mathbb{C}^n, n\geq 2$, and $K$ be a compact subset of $\Omega$ such that $\Omega\setminus K$ is connected.
From Hartogs' extension theorem, we know each holomorphic function over $\Omega\setminus K$ can be extended to a holomorphic function over $\Omega$. Does this extension is unique?
Hence, we need an extra assumption that $\Omega$ is connected to assure this extension is unique.
Is it necessary?
Hartogs' extension is unique, we just need to prove the following:
Theorem Let $\Omega$ be an open subset of $\mathbb{R}^n,n\geq 1$, $K$ be a compact subset of $\Omega$ and $\Omega\setminus K$ is connected, then $\Omega$ must be connected.
Proof If $\partial\Omega=\emptyset$, then $\Omega=\mathbb{R}^n$ is connected. Suppose $\partial\Omega\neq\emptyset$, and let $O_i, i\in I, |I|\geq 2$ be all connected components of $\Omega$, and $$K_i=K\cap O_i, U_i=O_i\setminus K_i.$$ Clearly, $U_i$ is open in $\Omega\setminus K$, and $$\Omega\setminus K=\cup_i U_i,\ U_i\cap U_j=\emptyset.$$ By the definition of connectedness, there must exist $j\in I$ such that $U_i=\emptyset$ for all $i\neq j$, then $O_i\subset K, \forall i\neq j,$ which implies $$0\leq d(K, \partial \Omega)\leq d(K,\partial O_i)=0, i\neq j,$$ contradicts with the fact that $$d(K,\partial \Omega)>0.$$
A counterexample: The closed topologist's sine curve.