Homotopy of a Continuous Proper Map to a Smooth Proper Map Working on a problem in Lee's Intro to Smooth Manifolds (2nd edition, problem 8, chapter 6).  The problem states:

Prove that a proper continuous map between smooth manifolds is
homotopic to a proper smooth map between said manifolds.

Let $M$ and $N$ be the manifolds with $F:M \to N$ a proper continuous map.  Since $F$ is continuous we automatically have the Whitney Approximation theorem: we can embed $N \subseteq \mathbb{R}^n$ for some $n$ and that $F$ is $\delta$-close to a smooth map $\widetilde{F}: M \to \mathbb{R}^n$.  Because $\widetilde{F}(M)$ can fit into some tubular neighborhood $U \subseteq \mathbb{R}^n$ of $N$, we have there is a mapping $r: U \to N$ which is a retraction and a submersion into $N$.  We can then take our homotopy to be $H: M\times [0,1] \to N$ given by
$$
H(x,t) \;\; =\;\; r \left ( (1-t) F(x) + t\widetilde{F}(x) \right ).
$$
The theorem shows that $F$ is homotopic to a smooth map $r\circ \widetilde{F}$, but we want to show that $r\circ\widetilde{F}$ is proper.
I'm somewhat lost as to proof strategies and need some help.  Lee hints that we need to show that $\tilde{F}$ is proper, and hopefully $r$ can easily show to be proper.  Where I'm getting stuck is in picking an arbitrary $K \subseteq N$ which is compact; how do we show $\widetilde{F}^{-1}(K)$ is compact as well?  I think we can pick a sequence in $\widetilde{F}^{-1}(K)$ and find a convergent subsequence.  Somewhere in there I'm assuming we need to use the fact that $||F(x) - \widetilde{F}(x)|| < \delta(x)$ for all $x \in M$, given $\delta:M \to \mathbb{R}_+$.
Any help is appreciated on this.  Thanks in advance!
 A: Recall Theorem 6.24 and Proposition 6.25: Let $M\hookrightarrow \Bbb R^n$ be an embedded smooth submanifold of dimension $m$. For each $x\in M$ define $N_xM:=\big\{v\in T_x\Bbb R^n\big| v\perp T_xM\big\}$ and $NM:=\big\{(x,v)\in \Bbb R^n\times \Bbb R^n:x\in M,v\in N_xM\big\}$.
Also, define $\pi\colon NM\to M$ as $(x,v)\longmapsto x$. Then, $NM$ is a $k$ dimensional smooth submanifold of $\Bbb R^n\times \Bbb R^n$ and $\pi$ is a submersion.
Define a map $E\colon NM\to \Bbb R^n$ as $(x,v)\longmapsto x+v$. Then at any point $(x,0)\in NM$ the map $dE$ is non-singular, i.e., for each $x\in M$ we have $\delta>0$ such that $E$ maps diffeomorphically $V_\delta(x):=\big\{(x',v')\in NM: |x-x'|<\delta,|v'|<\delta\big\}$ onto an open neighborhood of $x$ in $\Bbb R^n$.
Now, let $$\rho(x):=\sup\big\{\delta\leq 1: E\text{ maps }V_\delta(x)\text{ diffeomorphically }$$$$\text{onto an open neighborhood of }x\text{ in }\Bbb R^n\big\}.$$ Then, $\rho\colon M\to (0,\infty)$ is continuous and $0<\rho \leq 1$.
Next, define $V:=\left\{(x,v)\in NM:|v|<\frac{1}{2}\rho(x)\right\}$. Then, $E$ maps diffeomorphically $V$ onto an open subset $U$ of $\Bbb R^n$ with $M\subseteq U$.
Now, define $r\colon U\to M$ by $r:=\pi\circ E^{-1}$. So, $r$ is a retraction and submersion.

Observation: Let $y,y'\in U$ with $|y-y'|<\epsilon$ for some $\epsilon>0$. Write, $E^{-1}(y)=(x,v)\in V$ and $E^{-1}(y')=(x',v')\in V$, i.e. $r(y)+v=x+v=y$ and $r(y')+v'=x'+v'=y'$. So, $$|x-x'|=\big|(y-y')-(v-v')\big|\leq |y-y'|+|v-v'|$$$$\leq \epsilon+|v|+|v'|\leq \epsilon+\frac{1}{2}\rho(x)+\frac{1}{2}\rho(x')\leq \epsilon+1$$ as $\rho\leq 1$. Therefore, $$|y-y'|<\epsilon\implies \big|r(y)-r(y')\big|\leq\epsilon+1.$$

$\textbf{Theorem:}$ Now, let $F\colon N\to M$ be a continuous proper map. Then, there is a proper map $H\colon N\times [0,1]\to M$ such that $H(-,0)=F$ and $H(-,1)\colon N\to M$ is smooth proper map.
$\textbf{Proof:}$ Let $r\colon U\to M$ be as given above. Let $\delta(x):=\sup\big\{\varepsilon \leq 1:B_\varepsilon(x)\subseteq U\big\}$ for each $x\in M$. Then, $\delta:M\to (0,\infty)$ is a continuous function using similar argument just like to show $\rho$ is continuous. Also, $\delta\leq 1$. Let $\widetilde \delta:=\delta\circ F$ and choose a smooth function $\widetilde F:N\to \Bbb R^n$ such that $\left|\widetilde F(y)-F(y)\right|<\widetilde\delta (y)$ for each $y\in M$. In particular, $\left|\widetilde F-F\right|\leq 1$. Define, $$H(p,t):=r\left((1-t)F(p)+t\widetilde F(p)\right)\text{ for }(p,t)\in N\times [0,1].$$
This is well-defined because $\left|\widetilde F(p)-F(p)\right|<\delta\big(F(p)\big)$ i.e. $\widetilde F(p)\in B_{\delta\big(F(p)\big)}\big(F(p)\big)\subseteq U$.
Now, for $(p,t)\in N\times [0,1]$ we have $$\left|(1-t)F(p)+t\widetilde F(p)-F(p)\right|\leq t\cdot\left|\widetilde F(p)-F(p)\right|\leq 1$$ $$\implies\big|H(p,t)-r\circ F(p)\big|=\big|H(p,t)-F(p)\big|\leq 2.$$
Now, consider a $\textbf{proper}$ embedding $N\hookrightarrow \mathbb R^k$ for some $k$. Note that any proper map from a topological space to a locally compact, Hausdorff space is a closed map. Hence, we can think $N$ as a closed subset of $\mathbb R^k$. With induced metric on $N$ from $\mathbb R^k$ and with usual distance metric on $[0,1]$, consider the product metric on $N\times [0,1]$. So, a closed and bounded subset of $N\times [0,1]\subseteq_\text{closed}\Bbb R^{k+1}$ is compact by Heine–Borel theorem. Similarly, considering the proper embedding $M\hookrightarrow \Bbb R^n$ and giving induced metric on $M$ from $\Bbb R^n$, we can say any closed-bounded subset of $M$ is compact.
So, to prove $H$ is proper, it is enough to show $H^{-1}(C)$ is closed and bounded in $N\times [0,1]$ for any compact set $C$ is $M$. Let $C$ be a compact subset of $M$. Now, $H^{-1}(C)$ is closed in $N\times [0,1]$ as $H$ is continuous and $C$ is closed in $M$. Hence, we only need to show $H^{-1}(C)$ is bounded subset of $N\times [0,1]$. Now, notice that $\big\{(p_n,t_n)\big\}\subseteq N\times [0,1]$ is bounded if and only if $\{p_n\}\subseteq N$ is bounded. Also, $F$ is proper implies $\big\{F(p_n)\big\}$ is bounded if and only if $\{p_n\}$ is bounded, for any sequence $\{p_n\}$  in $N$. Finally, note that $\big| H(p,t)-F(p)\big|\leq 2$ for all $(p,t)\in N\times [0,1]$. So, if there were an unbounded sequence $\big\{(p_n,t_n)\big\}\subseteq  H^{-1}(C)$, then $\{p_n\}$, and hence $\big\{F(p_n)\big\}$ would be unbounded, a contradiction, as $C$ is a bounded subset of $\Bbb R^n$.  So, $ H$ is a proper map. $\square$
