If space $X$ deformation retracts to a point $x\in X$, then for each open $U\in X$ containing $x$ there exists an open $V\in U$ again containing $x$ s.t. inclusion of $V$ into $U$ is nullhomotopic.
My attempt: Since $X$ deformation retracts to a point $x$, there is a corresponding map $F:X\times I \to X$. So restriction of this map could be a solution. I suppose, the restriction of this maps domain to the set $V=pr_{X}\left(F^{-1}(U)\right)$ would be the right choice.
My geometrical understanding of this problem is next: We have to find a neighbourhood $V$ of $x$ s.t. at any time $t$ its points do not "get out" from the neighbourhood $U$. In other words the $F(V\times I) \subset U$.
Does my idea make sense? Would you correct anything if it does?