An Uncountable Subset of a Topological Space with a Countable Base $\Rightarrow$ M contains a limit point

Question

Please provide hints and not a direct answer as I would like to figure this out myself.

Prove that if $M$ is an uncountable subset of a topological space with a countable base, then some point of $M$ is a limit point of $M$

So far, If M has a countable base then there exits a countable everywhere dense subset $N \subset M$ such that $[N]=M$ $\Rightarrow$ $M$ contains all the limit points of $N$. The only way for our claim to be false is if N is made entirely of interior points $\Rightarrow$ N is open. Then $M$ is open is this a contradiction? Any hints on how to proceed would be greatly welcomed.

Answer 1

Sketch: If $M$ doesn't have a limit point, then for each $x\in M$ take an open set $U_x$ with $U_x\cap M\setminus\{x\}=\emptyset$. Then there is $B_x\in\mathcal B$ with $x\in B_x\subset U_x$, by the definition of a basis.

But $x\ne y\implies B_x\ne B_y$. This implies $M$ is countable. Contradiction.