Let $A \subset X$ be a subset, where $I =[0,1]$.
I'd like to prove a suggestion that for any retraction $r : X\times I \to X\times \{0\} \cup A\times I$, if some $x\in X$ satisfies the condition $t=P_I \circ r (x,t)$ for all $t \in I$, where $P_I $ is the projection on $I$, then $x \in A$.
It suffices to consider for the case $x\in \bar{A} \setminus A$.
$X$ is a topological space not necessary to be Hausdorff.
How to prove this?
Thanks in advance.
EDIT
For clarity, I edited my question.
Assume that $\bar{A}\setminus A \neq \emptyset$.
This question arises from Theorem 1.5 on page 431 in Bredon's Topology and Geometry.
Let $X = \{0, 1\}$ with trivial indiscrete topology, let $A = \{1\}$, and let $r: X × I → X × \{0\} ∪ A × I$ be defined by $r(x, t) = (1, t)$ for $t > 0$ and $r(x, 0) = (x, 0)$. Then $r$ is a retraction. It is continuous since $P_I ∘ r = P_I$ and $P_X ∘ r$ is continuous trivially. Therefore, $r$ is a counterexample to your claim.