Number of distinct points in $A$ is uncountable How can one show:

Let $X$ be a metric space and $A$ is subset of $X$ be a connected set with
  at least two distinct points then the number of distinct points in $A$
  is uncountable.


We will show that if $A$ is countable then $A$ is not connected. 
Let $a,b$ be two distinct points in $A$ and let $d$ be the metic on $X$. Then, since $d$ is real valued, there are uncountably many $r\in \mathbb R$ such that $0<r<d(a,b)$. Let $r_0$ be such that $\forall x\in A$, $d(a,x)\ne r_0$ and $0<r_0<d(a,b)$. This is possible because we are assuming $A$ is countable. Then the two open sets $U$, $V$ defined by $$U=\{x\in X: d(a,x)<r_0\}\cap A$$ and $$V=\{x\in X: d(a,x)>r_0\}\cap A$$ are disjoint, their union equals $A$ and $U\cap \bar V=\bar U\cap V=\emptyset$.
Therefore, $A$ is not connected.
