Milnor's proof that $f^{-1}(y)$ is a finite set

Question

Suppose $f : M \to N$ is a smooth mapping between manifolds of the same dimension $m$.Milnor says the following

Observe that if $M$ is compact and $y \in N$ is a regular value, then $f^{-1}(y)$ is a finite set (possibly empty) For $f^{-1}(y)$ is in any case compact, being a closed subset of a compact space $M$; and $f^{-1}(y)$ is discrete, since $f$ is one-to-one in a neighbourhood of each $x \in f^{-1}(y)$

The fact that $f^{-1}(y)$ is closed in $M$ follows from the fact that $\{y\}$ is closed in $N$ and since $f$ is continuous, we have $f^{-1}(y)$ to also be closed

But I can't see how the fact that $f$ is one-to-one in a neighbourhood of each $x \in f^{-1}(y)$ implies that $f^{-1}(y)$ is finite. The fact that $f$ is one-to-one follows from the inverse function theorem

Answer 1

If the set were not finite it would have an accumulation point, since $M$ is compact. Now in that point you also have $f(p) = y$ by continuity, but the injectivity conclusion of the inverse function theorem for a neighbourhood of that point would fail (by construction).

Answer 2

Since $y$ is a regular value, then you know that $f^{-1}(y)$ is going to be a 0-manifold, hence each point in $f^{-1}(y)$ has a neighborhood that is homeomorphic to a point (i.e. a neighborhood about each point containing only that point). But by compactness of $M$ if $f^{-1}(y)$ is infinite it contains an accumulation point, hence such a neighborhood could not exist at that accumulation point, violating your usual structure theorem for inverse images of regular values.