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 $\mathrm{E}^\eta.$

Let $x,y\in E^\eta$. Then there exist $x_E,y_E\in E$ such that $\|x-x_E\|$ and $\|y-y_E\|$ are lesser than $\eta$. Now, set $t\in(0,1)$. Then $tx_E+(1-t)y_E\in E$ and
$$\|tx+(1-t)y-tx_E-(1-t)y_E\|\le t\|x-x_E\|+(1-t)\|y-y_E\|<t\eta+(1-t)\eta=\eta$$
So $tx+(1-t)y\in E^\eta$. This proves the convexity of $E^\eta$.
Perhaps another proof: This is only an unpolished idea.
$E$ is an intersection of half-spaces: $E=\bigcap_i H_i$. Translate the border of each $H_i$ a distance $\eta$ orthogonally, moving away from $E$. You get new open half-spaces $H_i^\eta$, that contain $E^\eta$. Then $E^\eta=\bigcap_i H_i^\eta$.
