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.
This is a nice idea, but unfortunately the argument doesn't work because you can't guarantee that your extension map $\overline{h_i} : M \to S^n$ is continuous.
To develop intuition for why this is, let's study a classic example of a compact, non-Hausdorff topological manifold: an $n$-sphere with a double point at the south pole.
This manifold can be viewed as $M = \left( (S^n \times \{ a\}) \sqcup (S^n \times \{ b \} \right) / \sim$, where $S^n \sqcup S^n$ is the disjoint union of two $n$-spheres, and where $\sim$ is the equivalence relation that identifies all pairs of corresponding points from the two spheres except for the pair of south poles. $$ (x, a) \sim (x', b) \iff x = x' \text{ and } x \neq x_{\text{SP}} $$
From now on, we'll simplify our notation. We'll use $x_{\text{SP}, a}$ and $x_{\text{SP}, b}$ to denote the two south poles on $M$, and we'll refer to all other points on $M$ using the notation that we would use for the corresponding points on $S^n$.
To see that $M$ is non-Hausdorff, simply observe that if $V_a$ is an open neighbourhood of $x_{\text{SP}, a}$ and $V_b$ is an open neighbourhood of $x_{\text{SP}, b}$, then $V_a$ and $V_b$ have non-empty intersection.
So let's apply your construction to this manifold $M$ and see what happens.
Consider the open subset $U =M \setminus \{ x_{\text{SP}, b}, x_{\text{NP}} \}$. This $U$ contains $x_{\text{SP}, a}$, as well as all points that are not poles.
There is a homeomorphism $h : U \to S^n \setminus \{ x_{\text{NP}} \}$ given by $$ h(x) = \begin{cases} x & \text{if } x \neq x_{\text{SP}, a} \\ x_{\text{SP}} & \text{if } x = x_{\text{SP}, a} \end{cases}.$$
If we apply your construction to this $h$, then we would construct an extension map $\overline{h} : M \to S^n$, which maps all points outside of $U$ to $x_{\text{NP}}$. It is easy to see that this $\overline{h}$ is defined by $$ \overline{h}(x) = \begin{cases} x & \text{if } x \notin \{ x_{\text{SP}, a}, x_{\text{SP}, b}, x_{\text{NP}}\} \\ x_{\text{SP}} & \text{if } x = x_{\text{SP}, a} \\ x_{\text{NP}} & \text{if } x = x_{\text{SP}, b} \\ x_{\text{NP}} & \text{if } x = x_{\text{NP}} \end{cases}.$$
Unfortunately, this $\overline{h} : M \to S^n $ is not continuous.
Indeed, if $\overline{h}$ were continuous, then continuity of $\overline{h}$ at $x_{\text{SP}, a}$ would imply that there exists an open neighbourhood $V_a$ of $x_{\text{SP}, a}$ that is mapped by $\overline{h}$ into the southern hemisphere of $S^n$, while continuity of $\overline{h}$ at $x_{\text{SP}, b}$ would imply that there exists an open neighbourhood $V_b$ of $x_{\text{SP}, b}$ that is mapped by $\overline{h}$ into the northern hemisphere of $S^n$. However, in view of a remark we made earlier, $V_a$ and $V_b$ must necessarily have non-empty intersection. Any point that lies in the intersection of $V_a$ and $V_b$ would be mapped by $\overline{h}$ into the northern hemisphere of $S^n$, and also into the southern hemisphere of $S^n$. This is a contradiction.