Space without isolated points I've been struggling with the following theorem which states:
If $(X,\tau)$ is a topological space such that $X=X'$, then for every $A\in\tau$ its true that $A\subseteq A'$.
My attempt goes as follows, suppose that there exists an $x\in A-A'$, then $A-\{x\}\neq\varnothing$, so that $V\cap(A-\{x\})=\varnothing$ for some $V\in\tau$.
Ive been stuck here for a very long time and I dont see where is the flaw, I would appreciate any help!

Assume the contrary, $A\nsubseteq A'$
Then, $\exists x\in A$ such that $x\notin A'$
Since, $x \notin A'$ , $\exists U\in \tau\space  $  such that $$U\cap A=\{x\}$$
Both $A, U\in \tau \implies A\cap U =\{x\}\in \tau$
Hence, $\exists \{x\}\in \tau$ which contain no points of $X$ other than $x$ and this implies $x\in X$ is an isolated point. (Contradiction! )
