Let $X$ be a topological space and $A \subset X$ be a closed subspace. Assume that $A$ is a strong deformation retract of $X$.
Is it true that for every open subspace $U \subset X$, $A \subset U$, there exists an open $V \subset U$, $A \subset V$ such that $A$ is a strong deformation retract of $V$?
What if we assume $X$ is a metric space?
I think the comb space is a counter-example (and is a metric space).
Observe that the comb space strongly deformation retracts onto the point $(0,1)$. First deformation retract the "spokes" $\{\frac{1}{n}\} \times [0,1]$ onto $\{\frac{1}{n}\} \times \{0\}$, then deformation retract $[0,1] \times \{0\}$ onto the origin, then deformation retract $\{0\} \times [0,1]$ onto the point $(0,1)$.
However, consider the intersection $U$ of the comb space with $[0,1] \times (\frac{1}{2}, 1]$. $U$ is open in the comb space, and any neighborhood of $(0,1)$ in $U$ intersects infinitely many of the spokes and hence is disconnected. A disconnected set cannot deformation retract onto a point, so we're done.