If $X$ is second countable and $A \subset X$ is uncountable, $A$ has a limit point.
Attempt:Suppose $A$ is uncountable and each point $a \in A$ is isolated. Let $\beta$ be a countable basis for $X$. Then for each $a \in A$ there is a $B_a \in \beta$ with $B_a \cap A=\{a\}$. Then $a \neq b \implies B_a \neq B_b$, and for each $x \in A$ there is a unique $B_x \in B$ containing $x$, with $B_x \cap A=\{x\}$. Thus $|\beta| \geq |A|$. But $A$ was assumed to be uncountable and $\beta$ countable, a contradiction, thus the assumption that $A$ has no limit points was false.
Comment: Could I get a proof verification if this attempt was completely correct? Thanks.
Best Answer
The "streamlined" version, without duplication etc. :
Let $\beta$ be a countable base for $X$ and suppose $A$ is uncountable and such that no point of $A$ is a limit point of $A$.
Then by definition for each $a \in A$ we have an open set, that we can take from $\beta$ (!), say $B_a \in \beta$, such that $B_a \cap A = \{a\}$. (We always have such an open set $O$ containing $a$, but then we pick $B_a \in \beta$ so that $a \in B_a \subseteq O$ of course, using the property of a base.)
The assignment $a \to B_a$ is injective from $A$ to $\beta$:
If $U_a =U_b$ then $\{a\} = U_a \cap A= U_b \cap A= \{b\} \to a = b$. So $\aleph_0 < |A| \le |\beta| = \aleph_0$, which is a contradiction.
So for an uncountable $A$ there must be some $a \in A$ that is a limit point of $A$.