Let $A \subseteq \mathbb{R}^n$ and let the kernel of $A$ be: $Ker(A) = \{x \in A\, |\,[x,y] \subset A, \forall y \in A\}$. I want to show that if $A$ is compact, then $Ker(A)$ is compact.
I even do not know where to start and how to start. Any help is appreciated in advance.
As a subset of $A$, $\ker(A)$ is obviously bounded. The hard part is to prove that it is also closed.
For this, let us assume $x_n\in\ker(A)$ and $x_n\to x$, and prove that $x\in\ker(A)$.
For every $y\in A$, we have $[x_n,y]\subset A$ i.e. $$\forall t\in[0,1],\;tx_n+(1-t)y\in A,$$ hence $$\forall t\in[0,1],\;tx+(1-t)y=\lim_{n\to\infty}tx_n+(1-t)y\in\bar A=A,$$ q.e.d.
Or more rapidly: $\ker A$ is closed, as the intersection of the closed subsets $F_{y,t}\subseteq\Bbb R^n$ defined, for $y\in A$ and $t\in(0,1]$, by: $$F_{y,t}=\frac1t A-\frac{1-t}ty.$$
Note that this theorem still holds in any normed vector space (even infinite dimensional), by the same proof(s) of closedness, and the Heine-Borel theorem replaced with: every closed subset of a compact space is compact (so, it suffices to prove that $\ker(A)$ is closed), and in a Hausdorff space, every compact subset is closed (so, we can use that $A$ is closed).
A subspace of $\mathbb{R^n}$ is compact iff it is closed and bounded by the euclidean metric d or the square metric $\rho$. Since A is compact in $\mathbb{R^n}$ it is bounded and closed. Any subspace of A must also be bounded since it is contained in A. So the only thing left is to show that Ker(A) is closed in the subspace topology of A.
