For this question, it is important that we do not assume a priori that manifolds are Hausdorff. By an $n$-dimensional manifold, I just mean a topological space which can be covered by open sets homeomorphic to $\mathbb{R}^n$.
If $M$ is a compact $n$-manifold, is $M$ automatically Hausdorff? I think this can be proven by embedding $M$ into Euclidean space, but I might be overlooking something. My reasoning is as follows.
Let $N\in S^n$ denote the North Pole. Cover $M$ with finitely many open sets $U_1, …, U_m$, each equipped with a homeomorphism $h_i : U_i \rightarrow S^n-\{N\}$. The map $\overline{h}_i: M \rightarrow S^n$ defined by
\begin{align}
\overline{h}_i (x)=\begin{cases}
h_i(x) &, x\in U_i\\
N &, x\notin U_i
\end{cases}
\end{align}
is a continuous extension of $h_i$. So we have a homeomorphic embedding $M \rightarrow (\mathbb{R}^{n+1})^m$ given by
\begin{align}
x \mapsto (\overline{h}_1(x), …, \overline{h}_m (x)).
\end{align}
Ergo, $M$ is Hausdorff.
Does this reasoning work? As far as I can tell, Hausdorffness was not used at any point. This came about because I have been wondering lately whether the second-countability and Hausdorff conditions can be dropped when restricting our attention to closed manifolds. Second-countability can definitely be dropped.
Best Answer
Your $\overline{h}_i$ is not necessarily an embedding. Consider $X=[0,1]$ and define $M=(X\times\{0,1\})/\sim$ where $(x,0)\sim (x,1)$ for all $x$ except $x=0$. In other words it is an interval with two origins.
This already gives an example of a compact non-Hausdoff manifold. And also shows why $\overline{h}_i$ is not an embedding (because given appropriate $U_i$, e.g. "upper" and "lower" interval, we have that preimage of $N$ is a point, and so $\overline{h}_i$ would be just a homeomorphism).