Proposition 4.8.3. Let $\seqi{X}$ be Hausdorff spaces, then $\prod_{i \in I}X_{i}$ is Hausdorff.
Proof. Let $x, y \in \prod_{i \in I}X_{i}$ with $x \ne y$, then there exists $i \in I$ such that $\pi_{i}(x) \ne \pi_{i}(y)$. In which case, there exists $U \in \cn(\pi_{i}(x))$ and $V \in \cn(\pi_{i}(y))$ with $U \cap V = \emptyset$. Thus $\pi_{i}^{-1}(U) \in \cn(x)$, $\pi_{i}^{-1}(V) \in \cn(y)$, and $\pi_{i}^{-1}(U) \cap \pi_{i}^{-1}(V) = \emptyset$.$\square$