The following is the Theorem 6.5.1 stated in Evans' PDE book, which is slightly modified so that the statement is self-explanatory.
Let $U \subset \mathbb{R}^{n}$ be an open bounded and connected subset. Consider an elliptic operator $$Lu = -\sum_{i, j=1}^{n}(a^{ij}(x)u_{x_{i}})_{x_{j}},$$ where $a^{ij} \in C^{\infty}(\overline{U})$, $a^{ij}=a^{ji}$, and it satisfies uniform ellipticity condition, i.e., there exists $\theta>0$ such that $$\sum_{i, j=1}^{n}a^{ij}(x)\xi_{i}\xi_{j} \ge \theta |\xi|^{2}$$ for a.e. $x\in U$ and for all $\xi \in \mathbb{R}^{n}$. Then the following holds:
(i) Each eigenvalue of $L$ is real.
(ii) Furthermore, if we repeat each eigenvalue according to its (finite) multiplicity, we have $$\Sigma = \{\lambda_{k}\}_{k=1}^{\infty},$$ where $$0<\lambda_{1} \le \lambda_{2} \le \lambda_{3} \le \cdots$$ and $\lambda_{k} \to \infty$ as $k \to \infty$.
(iii) Finally, there exists an orthonormal basis $\{w_{k}\}_{k \ge 1}$ of $L^{2}(U)$, where $w_{k} \in H_{0}^{1}(U)$ is an eigenfunction corresponding to $\lambda_{k}$: $$\begin{cases} Lw_{k}=\lambda_{k}w_{k}&\text{in }U, \\ w_{k}=0 & \text{on }\partial U,\end{cases}$$ for $k=1, 2, \ldots$.
In this theorem, do we need the connectivity of $U$? In the book, the connectivity of $U$ is assumed in whole section, but I think the proof does not uses this assumption at all. I'm curious if I am missing something.
Thank you.