Every manifold is locally compact? 
Theorem. Every Manifold is locally compact.

This is a problem in Spivak's Differential Geometry.
However, don't know how to prove it. It gives no hints and I don't know if there is so stupidly easy way or it's really complex. 
I good example is the fact that Heine Borel Theorem, I would have no clue on how to prove it if I didn't see the proof. 
So can someone give me hints. I suppose if it's local, then does this imply that it's homeomorphic to some bounded subset of a Euclidean Space?
 A: Manifolds are locally Euclidean, and Euclidean space is locally compact. Hence manifolds are locally compact.
A: By definition, if $X$ is a manifold, then every point $x \in X$ admits an open neighborhood $U$ which is homeomorphic to $\mathbb{R}^n$ ($n$ is allowed to depend on $x$).  Let $f: U \rightarrow \mathbb{R}^n$ be such a homeomorphism.  Let $B$ be a closed ball of finite radius about $f(x)$ in $\mathbb{R}^n$.  By Heine-Borel, $B$ is compact, hence so is its homeomorphic preimage $f^{-1}(B)$, which is therefore a compact neighborhood of $x$.
Almost the same argument shows that $X$ has even a neighborhood base of compact sets at every point, which for non-Hausdorff spaces, is a priori stronger than having a single compact neighborhood at any point.  In my opinion "locally compact" should mean this stronger condition.  (On the other hand, in my terminology, both "manifold" and "locally compact" include the Hausdorff condition.)
A: Well, recall that manifolds locally look like euclidean space, by "look like", I mean locally homeomorphic to. Therefore, topological properties in the local sense should be translatable from the euclidean sense to the manifold sense. I shall write a proof keeping this intuition in mind.
Recall that a topological space is locally compact if each point $p$ in $M$ has a neighborhood $U$ that is contained in a compact set.
Proposition: If $M$ is a topological manifold, then $M$ is locally compact.
Proof:
Let $p\in M$. Let $(U,\phi)$ be a chart about $p$ where $\phi:U\rightarrow \phi(U)$ is a homeomorphism and $\phi(U)$ is an open subset of $\mathbb{R}^n$ for some $n$. Let $r>0$ be a real number such that $B=B(\phi(p),r)\subseteq \phi(U)$. Shrinking $r>0$ if necessary, we may assume that $\overline{B}=\overline{B(\phi(p),r))} \subseteq \phi(U)$. Hence, $p\in \phi^{-1}(B)\subseteq \phi^{-1}(\overline{B})\subseteq U$. Therefore we are done.
A: I do not think the above answers are completely right, since the "Hausdorff" condition in the definition of topological manifolds must be needed.
The key is to prove the following: If $V\subset U\subset X$, and X is Hausdorff, $\bar{V}_{U}$ is compact. Then $\bar{V}_{U}=\bar{V}$.
Proof: By definition, we only need to show that $\bar{V}\subset \bar{V}_U$. Suppose $x\in\bar{V}$, if $x\notin \bar{V}_U$. Since X is Hausdorff and $\bar{V}_{U}$ is compact, $\bar{V}_{U}$ is closed in X. So $ (\bar{V}_{U})^{c}$ is open and $x\in (\bar{V}_{U})^{c}$. Since $x\in \bar{V}$, we must have $V\cap (\bar{V}_{U})^{c}$ is not empty. It is impossible since $\bar{V}_{U}$ contains $V$.
