Given a convex set in a normed vector space, take a neighbourhood of it. Is still convex? Consider a normed vector space and a set there, call it $\mathrm{E}.$ Define the neighbourhood $\mathrm{E}^\eta$ of $\mathrm{E}$ with radius $\eta > 0$ as the set of vectors $v$ whose separation from $\mathrm{E}$ differs less than $\eta;$ in symbols, set the function $$v \mapsto \rho(v, \mathrm{E}) = \inf_{e \in \mathrm{E}} \|v - e\|,$$ which is uniformly continuous since $|\rho(v_1, \mathrm{E}) - \rho(v_2, \mathrm{E})| \leq \|v_1 - v_2\|,$ and the neighbourhood is defined to be $\mathrm{E}^\eta = \rho( \cdot, \mathrm{E} )^{-1}(-\infty, \eta)$ (it is readily seen as an open set).
If $\mathrm{E}$ is convex, is $\mathrm{E}^\eta$ convex?
I think I (almost) got a proof but is rather messy; one take any two points in $\mathrm{E}$ and then consider open balls of radii $\eta$ centred at them and show that any segment commencing in one ball and terminating in the other one will remain in $\mathrm{E}^\eta;$ call $\mathrm{F}$ the union of all this possible segments. The $\mathrm{F}$ ought to be convex and coincidential with $\mathrm{E}^\eta.$

You can see that $E^\eta = \cup_{x\in E} B(x,\eta)$. Take two points $y_i\in E^\eta$ and their corresponding points $x_i\in E$ such that $y_i \in B(x_i,\eta)$. Let $X\subset E$ be a segment joining points $x_i$ and $Y$ be a segment joining points $y_i$. Our goal is to show that $Y\subset E^\eta$.
Let $y\in Y$, then there exist $a \in [0,1]$ such that $y=ay_1 +(1-a)y_2$. Now consider $E\ni x=ax_1 +(1-a)x_2$:
$$\|y-x\| =\|(ay_1 +(1-a)y_2) - (ax_1 +(1-a)x_2)\|\le a \|y_1-x_1\| + (1-a)\|y_2-x_2\|=\eta,$$
hence $y\in E^\eta$.
