What is the maximum number of points that a subspace $X \subset \mathbb{R}²$ can have so that $\mathbb{R}²$ induces the discrete metric in $X$? I could not understand this question on my textbook. It's in the chapter of Metric Spaces, and I do think it's refering to a metric induced by a function $f$.
What is the maximum number of points that a subspace $X \subset \mathbb{R}^2$ can have so that $\mathbb{R}^2$ induces the discrete metric in $X$?
 A: Take three points in $A,B,C\in\mathbb{R}^2$ such that the distance between any two of them is $1$. For instance, you can take $A=(0,0)$, $B=(1,0)$, and $C=\left(\frac12,\frac{\sqrt3}2\right)$. Could there be a fourth point $D\in\mathbb{R}^2$ whose distance to the other three is also $1$? No, because then it would belong to the circle centered at $A$ with radius $1$ and also to the circle centered at $B$ with radius $1$. There are only two points at the intersection of these circles. One of them is $C$ and the distance from the other one to $C$ is $\sqrt3$, which is not $1$.
Therefore, the answer is three.
A: Suppose we have such an $X$. Note that $\mathbb{R}^2$ has a countable base, so every subspace too. And a discrete second countable space has at most countably many points as all sets $\{x\}$ must be in any base for $X$.
Directly: for every $x \in X$ pick rationals $q_1,q_2,r_1, r_2$ such that $\{x\} = ((q_1,q_2) \times (r_1, r_2)) \cap X$, which can be done as $\{x\}$ is open in the subspace. Then this defines an injection of $X$ into the countable set $\mathbb{Q}^4$. 
As David Hartley noted we want the discrete metric on $X$ in which case the answer is $3$, a equilateral triangle. We cannot have $4$ by geometric reasons. (We can have a regular tetrahedron in $\mathbb{R}^3$ of course). 
