Extension of a function by being homotopic to extendable function I am currently studying a paper of Olszewski in which the following Lemma appears.

Let $K$ be a CW complex, $Z$ a separable metrizable space, and $X$ a subspace of $Z$ such that $K\in AE(X)$. Then every mapping $g:A\rightarrow K$ from a closed subset $C\subseteq Z$ extends over an open set $U\subseteq Z$ such that $C\cup X\subseteq U$.

I'll include the short proof here. My question is about the last line.

Extend $g$ to a map $\hat{g}$ over the closure of an open neighbourhood $V$ of $C$ in $Z$. Then extend $\hat{g}\mid(\partial V\cap X)$ to a map $h$ over $X$. The map $f:V\cup X\rightarrow K$ defined by $f(x)=\hat{g}(x)$ for $x\in V$ and $f(x)=h(x)$ for $x\in X\setminus V$ is a well defined continuous map. By the Walsh Lemma $f$ is homotopic to a map $\hat{f}$ that can be extended to an open set $U\subseteq Z$ such that $C\cup X\subseteq V\cup X\subseteq U$. Thus $g$ is extendable in the desired way.

My issue is that I don't see why our function $f$ being homotopic to a map extendable over an open set would at all imply that is $g$ is extendable over an open set without first passing to some function homotopic to $g$. I assume I'm missing some basic result in homotopy theory? The Walsh Lemma doesn't seem to contain any content that would resolve the issue although I can include it if it would help illuminate anything. Any help would be appreciated.
 A: The proof in the paper is not very clear. Here is the trick:

Lemma (Borsuk): Suppose that $X$ is a normal space and $A\subseteq X$ a closed subset. Then for every neighbourhood $V$ of $A\times I\cup X\times 0$ in $X\times I$, there is a map $\varphi:X\times I\rightarrow V$ which is the identity on $A\times I\cup X\times 0$.

Proof: This is Lemma (8.2) in Borsuk's book The Theory of Retracts. As it is short I will give the proof.
For each $x\in A$ let $U_x\subseteq X$ be an open neighbourhood satisfying $U_x\times I\subseteq V$ (this uses the Tube Lemma). The set $U=\bigcup_{x\in A}U_x$ is then an open neighbourhood of $A$ with $U\times I\subseteq V$. Next, use the Urysoshn Lemma to find a map $\alpha:X\rightarrow I$ with $\alpha(X\setminus U)=0$ and $\alpha(A)=1$. Finally put $\varphi(x,t)=(x,\alpha(x)t)$. $\quad\blacksquare$
This is used as follows.

Lemma: Let $X$ be a normal space and $K$ a space. Suppose that $K\in ANE(X\times I)$. Then any closed $A\subset X$ has the homotopy extension property with respect to $K$.

Proof: Take a map $f:X\rightarrow K$ and a homotopy $F:A\times I\rightarrow K$, defined for a closed $A\subseteq X$ and satisfying $F_0=f|_A$. Because $A\times I\cup X\times 0$ is closed in $X\times I$, the map $F\cup f:A\times I\cup X\times 0\rightarrow K$ extends to $\tilde F:V\rightarrow K$, where $V$ is some neighbourhood of $A\times I\cup X\times 0$. Let $\varphi:X\times I\rightarrow V$ be the map supplied by Borsuk's Lemma and put $G=\widetilde F\circ\varphi:X\times I\rightarrow K$ to obtain the required homotopy. $\quad\blacksquare$
Returning to the main problem, I will the notation from Olszewski's Lemma 2: $K$ is a CW complex, $Z$ a separable metrisable space with closed subspace $C$, and $g:C\rightarrow K$ is a map. In the lemma a homotopy $h|_C\simeq g$ has been found, where $h:U\rightarrow K$ is a function defined on an open set $U\subseteq Z$ containing $C\cup X$ ($X$ being a certain subset of $Z$).
Now, the set $U$ is itself a separable metric space, and hence so is $U\times I$. Because of Kodama's Theorem (recanted by Olszewski in his introduction), we therefore have that $K\in ANE(U\times I)$. Since $C$ is closed $U$, by the Lemma above, it has the homotopy extension property wrt to $K$. Therefore the homotopy $h|_C\simeq g$ can be extented to a homotopy $G:U\times I\rightarrow K$ with $G_0=h$. The map $\widetilde g=G_1$ is an extension of $g$ over $U$.
