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First I give names to the maps by $$ 0\to A \xrightarrow{f} B \xrightarrow{g} C \to 0. $$ If $A$ is open in $B$, then consider $$ B = \coprod_{[x] \in B/A } x.A .$$ Now each $x.A$ is open so this is an open cover of $B$. But this means $C$ is discrete and compact, so $C$ is finite. So $B$ is covered by a finite number of open compact sets thus compact. ...


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As you said, you need to prove there is an $N\in\mathbb{N}$ such that if $n\ge N$ then $d(f_n(x),f(x))<\epsilon$ for all $x\in K$. In other words, you want $f_n$ to be in $B^K_\varepsilon(f) = \{g:X\to Y \mid d(g(x),f(x))<\varepsilon \text{ for all } x\in K \}$. So if you find pairs $(K_1,V_1),\dots,(K_l,V_l)$ such that $f\in W=\bigcap_{i=1}^l(K_i,V_i)$...


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To answer your additional question: Without assumptions on the boundary the statement is false. You can pretty much do the same thing as in the unbounded case. More precisely: look at the bounded part of a spiral (with some thickness) with infinite length. Parameterized along the length of the spiral from the outside in, just let $f_n=0$ on $[0,n]$ and $f_n =...



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