Suppose $Z$ deformation retracts onto a point $z \in Z$, i.e. the constant map $c_z: Z \rightarrow Z$ is homotopic to $id_Z$, where the homotopy is not necessarily relative to $\{z\}$.
Let $H$ be the homotopy from $c_z$ to $id_Z$.
I would like to show that given an open neighbourhood $U$ of $z$, I can choose an open neighbourhood $V$ in $U$ of $z$ such that the inclusion of $V$ into $U$ is homotopic to $c_z$ restricted to $V$ relative to $\{z\}$.
In the case where $H$ is a strong deformation retract and $Z$ is compact, this is straightforward: since $Z-U$ is closed, it's preimage under $H(Z,t)$ is closed, so there is some $t < 1$ such that $H(Z,t')$ is contained in $U$ for all $t' > t$. Let $V$ be an open subset of $H(Z,t')$, then let $H'(x,t') = H(x, t+ t'(1-t))$. Then $H'$ is a strong deformation retract of $V$ onto $\{z\}$.
I do not understand how to generalise this. It feels as though taking the path component of $z$ in $U$ may be helpful, but I don't understand how to actually show that this works.
Counter example:
Let $\mathbf{Q}\subseteq \mathbf{R}$ be equiped with the subspace topology. Put $Z = \mathbf{Q}\times[0, 1] / \mathbf{Q}\times\{1\}$ and $z_0 = (0, 0)\in Z$.
Now $z_0$ is a deformation retract of $Z$. It's clear that we can first deform $\mathrm{Id}_Z$ to the inclusion of the point $\mathbf{Q}\times\{1\}$ and then to the inclusion of $z_0$.
However, $U = \mathbf{Q}\times [0, 1[$ is a neighbourhood of $z_0$, and that $z_0$ has no neighbourhood $V\subseteq U$ such that the inclusion $V\hookrightarrow U$ is homotopic to $c_{z_0}$.