Prove that ANE space also has HEP 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.
 A: I believe I have found an argument that works.
Let $U$ be an open set such that $(X\times\lbrace 0\rbrace)\cup (A\times[0,1])\subseteq U$. Define $\mathcal{U}=\lbrace V\times W|\hbox{ }V \textrm{ is open in }X,\hbox{ } W \textrm{ is open in }[0,1] \textrm{ and }V\times W\subseteq U\rbrace$, the set of all boxes contained in $U$.
Lemma. For every $a\in A$ there exists a downward closed open neighborhood $U_a$ of $\lbrace a\rbrace\times[0,1]$ such that $U_a\subseteq U$.
(Here of course, a set $U'\subseteq X\times[0,1]$ is defined to be downward closed if for every $(x,t_1)\in U'$ and every $t_2\in[0,t_1]$ we have $(x,t_2)\in U'$.)
Proof. Let $a\in A$. Then $\lbrace a\rbrace\times[0,1]$ is compact, so there exist $V_1\times W_1,V_2\times W_2,...,V_n\times W_n$ in $\mathcal{U}$ such that $\lbrace a\rbrace\times[0,1]\subseteq\bigcup_{i=1}^n V_i\times W_i$. But $V = \bigcap_{i=1}^n V_i$ is a finite intersection of open sets that all contain $a$ (or at least we can safely remove those that do not contain $a$ and still get an open cover), so it must itself be an open set that contains $a$. Thus we see that $V\times\bigcup_{i=1}^n W_i = V\times[0,1]$ is an element of $\mathcal{U}$ and downward closed which proves the lemma.
Next, define $\mathcal{L}=\lbrace V\times W\in \mathcal{U}|\hbox{ }\exists t\in[0,1]:W=[0,t)\rbrace$ and $U_\infty=\bigcup\mathcal{L}$.
Observe that an arbitrary union of downward closed sets is again downward closed. Therefore $U_\infty$ is a downward closed set that contains $X\times\lbrace0\rbrace$.
Now just take $U'' = U_\infty\cup\bigcup_{a\in A} U_a$. This is a union of downward closed sets and therefore downward closed. It is contained in $U$ because every set in the union is contained in $U$. It contains $X\times\lbrace 0\rbrace$ and $\lbrace a \rbrace\times[0,1]$ for every $a\in A$. So it also contains $(X\times\lbrace 0\rbrace)\cup (A\times[0,1])$ and the problem is solved.
Any corrections/comments are welcome.
Edit: for better readability I shall state the conclusions here. The above argument hopefully shows that every open neighborhood of $F = (X\times\lbrace 0\rbrace)\cup (A\times[0,1])$ contains a downward closed open neighborhood of $F$. Since the extension $h':U\to Y$ of $h: F\to Y$ is continuous, so is its restriction to this smaller neighborhood $h'|_{U''}:U''\to Y$. Therefore this is also an extension of $h$. Since now $U''$ is downward closed, $H$ will now be well defined.
