Example of a convex set whose closure is not convex? An enumeration $ν\colon ℕ → A$ of the rationals $A$ in $(0..1)$ yields an open set $U_ν = \bigcup_{k ∈ ℕ} B_{1/4^k}(ν(k))$, containing all of $A$. You can choose $ν$ such that $U_ν ⊂ (0..1)$ (by using choice, I guess). Its Lebesgue measure is
$$λ(U_ν) ≤ \sum_{k = 1}^∞ 1/{4^k} = \frac 1 {1-1/4} - 1 = 1/3 < 1,$$
so $U_ν \subsetneq (0..1)$.
As $A ⊂ U_ν$ is dense in $[0..1]$, one ends up with a non-convex set $U_ν$ whose closure is convex.

But how about the other way around? Is there a convex set (in euclidian space/general space) whose closure fails to be convex? Furthermore, is there such an open set?

Also, is there a simpler example of a non-convex open set whose closure is convex?

The closure of a convex set is convex. Indeed, take $a,b\in \overline{A}$ and $t\in (0,1)$. For every $r>0$ we have $a',b'\in A$ such that $|a-a'|<r$ and $|b-b'|<r$. The triangle inequality implies
$$
|(ta+(1-t)b)-(ta'+(1-t)b')| < r
$$
Convexity of $A$ implies $ta'+(1-t)b'\in A$. Since $r$ was arbitrary, $ta+(1-t)b\in \overline{A}$.
