Proof. Let $x\in E^{c}$. For each $y\in E$, there exists $U_{y}\in\cali{N}(y)$ and $V_{y}\in\cali{N}(x)$ such that $U_{y}\cap V_{y}=\emp$. By compactness, there exists $E_{0}\suf E$ finite such that $\cups{}{y\in E_0}U_{y}\supf E$. In which case $E\cap \caps{}{y\in E_0}V_{y}=\emp$ and $\caps{}{y\in E_0}V_{y}\in \cali{N}(x)$. Thus $E^{c}$ is open.$\square$