I have the following definition of a neighbourhood deformation retract (or NDR, for short): A pair $(X, A)$ is called an NDR if $A\subseteq X$ is closed and there exists a neighbourhood $A\subseteq V\subseteq X$ and a retraction $r: V\rightarrow A$, i.e. $r|_A=\mathrm{id}_A$ and there is a homotopy $H: V\times [0, 1]\rightarrow V$ such that $H(-, 0)=r, H(-, 1)=\mathrm{id}_V$ and $H|_{A\times [0, 1]}=\mathrm{pr}_A$.
Now I was wondering about the following: If $A\subseteq B$ is an NDR and $B\subseteq X$ is an NDR, does it follow that $A\subseteq X$ is an NDR? I suspect this is true, but somehow I cannot produce a proof.
I was trying to argue as follows: There is a neighbourhood $A\subseteq V\subseteq B$ and a retraction $r: V\rightarrow A$; moreover, there is a neighbourhood $B\subseteq W\subseteq X$ and a retraction $s: W\rightarrow B$. Now I want to choose $V'=s^{-1}(V)\subseteq X$ as a neighbourhood of $A$ in $X$ and use $r\circ s: V'\rightarrow A$ as a retraction. However, the problem is that I do not know whether the homotopy $H_s: W\times [0, 1]\rightarrow W$ between $s$ and $\mathrm{id}_W$ restricts to a homotopy $V'\times [0, 1]\rightarrow V'$, i.e. if it keeps points of $V'$ inside $V'$. Am I missing something here?