Find a distance function $d$ s.t a set is closed wrt $d \Leftrightarrow$ it is a finite union of subspaces Expanding the compressed title, the question is:
Is there a distance function $d$ on $\mathbb{R}^2$ such that a set is closed with respect to $d$ if and only if it is a finite union of affine subspaces of $\mathbb{R}^2$?
So I know that an affine subspace of $V$ over a field $k$ is a set of the form $v+U=   \left \{ v+u : u \in U \right \} $ where $ v \in V $ and $  U⊂V$ is a linear subspace.
Breaking down this question into formal maths and applying it to our case we can re-write this as:
Does a distance function $d$ on $\mathbb{R}^2$ exist s.t $\exists$ a set $S$ s.t $\forall x \in \overline{S} \space\exists \space \epsilon > 0 $ s.t $B(x,\epsilon) \subseteq \overline{S} \iff S = \bigcup_{n<N} \left \{ v+u: u \in U \right \} $ where $ v \in \mathbb{R}^2, U \subseteq \mathbb{R}^2, N \in \mathbb{N}$ ?
Am I doing this right? Could someone help me get to the answer. I'm a bit overwhelmed. 
Thanks.
 A: Suppose so. Then the space is in particular not discrete (or else every subset would be closed), so pick some sequence $x_n$ that converges to a point not in $x_n$. Call the limit $x$. First, the $x_n$ accumulate to no other point (that is, the closure of $\{x_1, x_2, \dots, \}$ adds only the point $x$; this is because if there was some subsequence $x_{n_k}$ and a point $y$ with $d(x_{n_k},y) < 1/k$, then $d(x,y) \leq d(x,x_{n_k}) + d(x_{n_k},y)$, which goes to $0$ as $k \to \infty$ because $x_n \to x$. So  $x=y$.
So the closure of the set $\{x_1, x_2, \dots, \}$ is the countable set $\{x_1, x_2, \dots, x\}$, but your space only has finite and uncountable closed subsets. Contradiction.
The same argument shows that any infinite metric space has a countable closed subset.
A: I hope you know that $\Bbb{R}^2$ is not a finite union of proper subspaces. From this follows that taking any two closed sets $C,D \neq \Bbb{R}^2$, you have $C \cup D \neq \Bbb{R}^2$ (in other words, the plane is not a finite union of lines or points).
Using the De Morgan law, this is equivalent on saying that taking any two non-empty open sets $U,V \neq \emptyset$, you have $U \cap V \neq \emptyset$.
But now, this cannot hold in a metric space (with at least two points): in fact, let $x,y \in \Bbb{R}^2$ be two distinct points, and call $3d = d(x,y) > 0$. Then by the triangle inequality the two open balls of radius $d$ centered at $x,y$ respectively are two disjoint non-empty open sets.
