This is a problem that I'm doing. Here $V \hookrightarrow U$ is the inclusion map of $V$ in $U$.
I don't understand the point of having $V \subset U$. In the following I'm presenting a solution that seems to suggest that any neighborhoods $U$ of $x$ would already have that property. Please correct me if I'm wrong, and please keep your answer elementary as I'm trying to build things up from the basic definitions.
Let $g$ be the retraction of X onto $\{x\}$. Since $X$ deformation retracts onto $\{x\}$, there exists a homotopy $H: X \times I \to X$ such that, for all $y \in X$, $H(y,0) = id_X(y) = y$, and $H(y,1) = i_{\{x\}}(g(y)) = i_{\{x\}}(x) = x$, where $i_{\{x\}}$ is the inclusion map $\{x\} \hookrightarrow X$.
Let $U$ be a neighborhood of $x$ and $i_U$ be the inclusion map $U \hookrightarrow X$. Then I define the homotopy between $i_U$ and the constant map that sends $p \in U$ to $x$ as follows: $F: U \times I \to U$, $F(p,t) = i_U^{-1}(H(i_U(p),t))$. Would this map be acceptable? If no, why not?
The question is if your $H$ is a deformation retract or a strong deformation retract, since if $H$ is not strong deformation retract it could happen that $F(i_U(p),t)) \not \in U$ for $t \neq 0,1$. Thus $i_U^{-1}(F(i_U(p),t)))$ and then also your $F$ might be not well defined.