Proof about fundamental class for M exists then M is compact. 
In this proof, I can't see why the claim"this image is compact" is true, and I don't understand why it implies $M$ is compact.

I have the same question on the remark of Hatcher's Theorem 3.26. Maybe the argument for $M$ being compact is as hatcher argues briefly: "Furthermore, $M$ must be be compact since $\mu_x$ can only be nonzero for $x$ in the image of a cycle representing $\mu$, and this image is compact":
Suppose $\mu = [c]$, where $c$ is a $n$-cycle in $S_n(M; R)$. Let $c=\sum_{i}r_i \sigma_i$. Then $\mu_x$ in $H_n(M|x; R)$ is the equivalent class $c$ representing in $S_n(M, M-x; R)$. $\mu_x$ is nonzero means that there is some $i$, such that $\sigma_i: \Delta^n \rightarrow M$ satisfying $x \in \sigma_i(\Delta^n)$. Otherwise $c$ is a singular $n$-chain in $S_n(M-x; R)$. Since for all $x \in M$, $\mu_x$ is nonzero, $\cup_i \sigma_i(\Delta^n) = M$. And for $\Delta^n$ being compact, $\sigma_i(\Delta^n)$ is compact for all $i$. And $M$ is the finitely union of compact subspaces. Hence $M$ is compact.
