Finding a nbhd of a point in image of a proper maps s.t.whose preimage can be contained in the other nbhd. Let $f:M\to N$ be a continuous and proper map.
Let $a$ be in $\operatorname{Range}(f)$, and $f^{-1}(a)$ contained in a nbhd $U$.
Why is there a nbhd $W$ of $a$ s.t. $f^{-1}(W)$ is contained in U?
 A: It’s true because proper maps (also sometimes known as perfect maps) are closed. Let $K=f^{-1}[\{a\}]$, and let $U$ be an open set containing $K$. Let $F=M\setminus U$; $F$ is closed, so $f[F]$ is closed in $N$. Let $W=N\setminus f[F]$; clearly $W$ is open in $N$, and it’s not hard to check that $a\in W$ and $f^{-1}[W]\subseteq U$.
Actually, there’s more than one definition of proper map. They’re equivalent for nice spaces, but not in general. In particular, if you define a map $f$ to be proper if $f^{-1}[K]$ is compact whenever $K$ is compact, then $f$ need not be closed, and your statement need not be true. For a counterexample, let $\mathscr{C}$ be the cofinite topology on $\Bbb{N}$, let $\mathscr{I}$ be the indiscrete topology on $\Bbb{N}$, and let $f$ be the identity map on $\Bbb{N}$. Every subset of $\Bbb{N}$ is compact in both topologies, so $f$ is perfect by this definition. However, $f$ is not closed, since $\{1\}$ is closed in $\langle\Bbb{N},\mathscr{C}\rangle$ but not in $\langle\Bbb{N},\mathscr{I}\rangle$. To see that it doesn’t satisfy your property, let $a=1$ and $U=\Bbb{N}\setminus\{2\}$: $U$ is an open nbhd of $f^{-1}[\{a\}]=\{1\}$ in $\langle\Bbb{N},\mathscr{C}\rangle$, but the only open sets in $\langle\Bbb{N},\mathscr{I}\rangle$ are $\varnothing$ and $\Bbb{N}$, and neither $f^{-1}[\varnothing]$ nor $f^{-1}[\Bbb{N}]$ is a subset of $U$.
