Let $Y$ be a normal space, we say that it is ANE (Absolute Neighborhood Extensor) if for every metric space $X$ and closed subset $A$ of $X$, if $h\colon A\to Y$ then there is an open neighborhood $U$ of $A$ that we can extend $h$ to h'\colon U\to Y.
We say that $Y$ has HEP (Homotopy Extension Property) if for every metric space $X$ and closed subset $A$ of $X$, if $h\colon(X\times\{0\})\cup(A\times[0,1])\to Y$ is continuous then we can extend it to $H\colon X\times[0,1]\to Y$.
I want to show that $Y$ is ANE then $Y$ has HEP. So let $Y$ be ANE and $X$ a metric space, $A\subseteq X$ closed.
Suppose $F=(X\times\{0\})\cup(A\times[0,1])$ and $h\colon F\to Y$ is continuous. We can therefore find an open $U\supseteq F$ and extend $h$ to h'\colon U\to Y. By Urysohn's lemma there is $g\colon X\times[0,1]\to[0,1]$ continuous such that: $g(x,t)=1$ for $(x,t)\in F$ and $g(x,t)=0$ for $(x,t)\notin U$.
Define H(x,t)=h'\Big(x,\ g(x,t)\cdot t\Big). Clearly $H$ extends $h$ continuously, however there is a fine detail which I cannot overcome:
Suppose $(x,t)\in U$ why does $(x,\ g(x,t)\cdot t)\in U$ as well? I cannot think of a reason why this should be true, but I also cannot find an argument why this must not be possible to have (in a sense, $U$ should be closed downwards in the second coordinate).
Any hints, tips or corrections to my suggested solution above are welcomed.