Condition for set to be closed. Consider the convex set with the following property: $A \subseteq\mathbb{R}^n$, and $A \cap [a,b]$ is closed for all segments. Then $A$ is also closed.
My idea is fully geometrical. Suppose we have some limit point $a$ that doesn't belong to $A$. Then $\forall \varepsilon > 0$ $B_{\varepsilon}(a) \cap A \ne \emptyset$. So there are at least two points $a_\varepsilon, b_\varepsilon \in A$ (otherwise $a$ isn't limit point). So we have $[a_\varepsilon, b_\varepsilon] \cap A = [a_\varepsilon, b_\varepsilon]$ (because of convexity if $a_\varepsilon, b_\varepsilon \in A$, then $\alpha a_\varepsilon + (1 - \alpha) b_\varepsilon \in A$). Let $\varepsilon \to 0$, we have a sequence of decreasing segments $[a_\varepsilon, b_\varepsilon] \to \{a\}$.  Since all intersection with segments are closed, then in limit $a$ will be in $A$.
N.B. to be added: these $[a_\varepsilon, b_\varepsilon]$ shouldn't be nested. That's the crucial point in this proof. Also from geometrical point of view if $a$ not in $A$, there should be a segment $[a, c]$ :  $[a,c] \cap A = [a, c]$ and hence $a \in A$, but I'm not certain how to prove this staitment.
Here I'm not sure about the last proposition: a will be in A. How can I show it more precisely?
 A: I do not see any easy way towards the conclusion from your arguments, you did not yet use the assumed property explicitly.
I propose a solution by "induction" with respect to the dimension. I will explain it for $d=2,3$, higher should also work this way, but the apparent large number of exceptional cases makes it much more tedious.
First $d=2$: we may assume that $a = (0,0)$ and there exists $a_1 = (1,0)\in A$. Then, since $a$ is a limit point, there exist $a_n = (x_n,y_n)\in A$ such that $|a_n|\leq \frac 1n$. We may assume in addition that $x_n\geq 0$, because the interval $[a_1,a_n]$ belongs to $A$, hence if there was a point on it whose $x$ coordinate vanishes, then it would be closer to 0 than $a_n$.
If there is an infinite number of $a_n$'s both over and under the $x$-axis, it is easy to see that the interval $(a,a_1]$ belongs to $A$, and so $a\in A$ by our assumption. Assume then that almost all $a_n$'s lie strictly above the $x$-axis. Consider triangles $\Delta a_1 a_2 a_n$ for $n\geq 3$ (almost all of them are non-degenerate). Their union is a subset of $A$ and it is easy to see that for every $x\in (0,1)$ it contains a line segment of the form $\{x\}\times (0,\epsilon(x))$. Thus, our assumption implies that $A$ contains every point $(x,0)$ for $x\in (0,1)$ and the assertion follows. This ends the case $d=2$.
For $d=3$ we start with similar simplifications: $a=0$, $a_1 = (1,0,0)$, and $a_n$'s have non-negative $x$ coordinate (i.e. $x_n\geq 0$). Assume (WLOG) that $y_2=0$, $z_2>0$ and let $\pi= OXY$ – the vertical plane. If there is an infinite number of $a_n$'s below $OXY$, then the situation reduces to $d=2$, because $a$ is then a limit point of the two-dimensional convex set $A\cap \pi$. We can also reduce to $2d$ if an infinite number of $a_n$'s belong to $OXZ$ or lie on both sides of $OXZ$. We have therefore reduced situation to $x_n,y_n,z_n> 0$ for $n\geq 3$. Consider the tetrahedra $\Delta a_1a_2a_3a_n$ for $n\geq 4$. Their union is a subset of $A$ and for all $(x,0,z)$ belonging to the triangle $\Delta aa_1a_2$ there exists an interval $[(x,0,z),(x,\epsilon(x,z),z)]$ belonging to $A$. Thus our assumption implies that $\Delta aa_1a_2\subseteq A$ and so $a\in A$.
