X is second countable then $A\subset X$ has an accumulation point I am having a pleasent friday with topology exercises and found myself kind of missing something on this proposition. It seems I`m not using - at least not explicitly- all the hypothesis provided. I will show you my attempt and would like to know wether it is right or wrong and, if it is right, where am I using the fact that A is non-countable.
Proposition: Let $(X, \tau)$ be a second-countable topological space and $A\subset X$ non-countable. Then, there`s an $x\in A$ such that $x$ is an accumulation point, i.e.,  
$$\exists x\in A:( \forall V_x \in \tau: x\in V_x\ \ \wedge \ \ V_x \backslash\{x\} \cap A\neq \emptyset) $$ 
Proof:
By hereditarity of second-countability  $A$ is second countable.
Second-countability implies separability
Hence, $A$ is a separable space, i.e., there is such a $D\subset A$ dense in $A$.
Let $x\in A \backslash D$ and $V_x \in \tau$ such that $x\in V_x$
Provided that $D$ is dense, every open set of $A$ has points of $D$
Hence $(V_x\backslash\{x\})\cap D$ has a point of $D$
Which is clearly not $x$ because I choosed $x\in A\backslash D$ and $A\backslash D$ is non-void as it is the difference of a uncountable set A and a countable set D.
From $(V_x\backslash\{x\})\cap D\subset (V_x\backslash\{x\})\cap A$  it follows that  $(V_x\backslash\{x\})\cap A\neq \emptyset$    $ \ \ \ \ \ \square$
 A: Your argument comes down to : $A$ is separable (any subset of second countable space is), so let $D \subset A$ be dense in $A$ and countable. Then any $x \in A \setminus D$ (which exists as $A$ is uncountable and $D$ is countable) is as required, as for any open $V$ containing $x$, $(V\setminus\{x\}) \cap A$ is open in $A$ and must intersect $D$, hence $A$. This is certainly correct. There are other ways, though:
Use the second countability directly:
Let $B_n$, $ n \in \mathbb{N}$ be a countable open base for $X$.
Suppose we have no such $x$. Then for each $x \in A$ we can find $V_x$ open and containing $x$, such that $V_x \cap A = \{x\}$. But this means we can can find $n(x)$ such that $x \in B_{n(x)} \cap A \subseteq V_x \cap A = \{x\}$.
But $A$ is uncountable and $\mathbb{N}$ is uncountable, so there are $x \neq y \in A$ such that $n(x) = n(y) = n$. But this is a contradiction because 
$$\{x\} = B_{n(x)} \cap A = B_n \cap A = B_{n(y)} \cap A = \{y\}$$ contradicting $x \neq y$. So we must have such an $x$ in $A$.
Use that $A$ is Lindelöf:
We can also use that $A$ is Lindelöf (every cover has a countable subcover) which also follows from second countability of the whole space. So starting by contradicition again, as before, with the $V_x$ containing $x$ with $V_x \cap A = \{x\}$, we note that $\{V_x \cap A: x \in A\}$ is an open cover of $A$, so countably many, say $V_x, x \in B$, where $B$ is countable, already cover $A$. But then 
$$A \subseteq \bigcup \{V_x: x \in B\} \cap A = \bigcup \{V_x \cap A: x \in B\} = \bigcup\{\{x\}, x \in B\} = B$$
contradiction as $B$ is countable and $A$ is uncountable.
Note that the above argument would also yielded the same contradiction if we merely had assumed all $V_x \cap A$ were countable (countable union of countable sets is countable). So in fact we can prove there must be some $x$ with $V \cap A$ uncountable for all open $V$ containing $x$. Using slightly less, gives us a bit more.
