Connected metric spaces with at least 2 points are uncountable.

Solution 1:

Let your points be $a$ and $b$.

Let $\lambda\in(0,1)$ Suppose there is no point $x$ in your space such that $d(a,x)=\lambda d(a,b)$. Then the sets $\{z:d(a,z)<\lambda d(a,b)\}$ and $\{z:d(a,z)>\lambda d(a,b)\}$ are two non-empty open sets which partition the space.

Since we are assuming connectedness, this is impossible.

Therefore, the image of the function $d(a,\mathord\cdot):X\to\mathbb R$ contains the interval $(0,d(a,b))$, and this can only happen if $X$ is uncountable.

Even if useless, this is a pretty result :)

Solution 2:

Call those two points $x_1$ and $x_2$. So $d(x_1,x_2)>0$.

What is the set $\{d(x_1,x) : x\in A\}$? Does it contain all numbers between $0$ and $d(x_1,x_2)$? If so, then you have at least as many points $x\in A$ as numbers between $0$ and $d(x_1,x_2)$. If not, then some number $c$ between $0$ and $d(x_1,x_2)$ is not the distance between any point $x\in A$ and $x_1$. So consider the two sets $$ \{x\in A : d(x_1,x)<c\}\text{ and }\{x\in A : d(x_1,x)>c\}. $$ Those are disjoint open subsets of $A$ whose union is all of $A$, so $A$ would not be connected.

Solution 3:

The biggest application that I know for this result is the immediate corollary that a countable metric space is totally disconnected.