I'll denote closure of A with $\overline{A}$, $A^\circ$ as the interior of A, $\partial A$ as the boundary of $A$ and $A'$ as the accumulation points. I'm trying to prove the following:
$$\overline{A}=A^\circ \cup\partial A$$
Is this a valid proof?
Take $x \in A^\circ \cup \partial A$ then $x \in A^\circ$ or $x \in \partial A$
if $x \in A^\circ$ then $x \in \overline{A}$
if $x \in \partial A$ then $x \in \overline{A} \cap\overline{(X\setminus A)}$ thus $x \in\overline{A} $ so $A^\circ\cup\partial A\subset\overline{A}$
Take $x \in \overline{A}$ then $x \in A' \cup A$ thus $x \in A'\setminus A$ or $x \in A^\circ$
if $x \in A'\setminus A$ then $x \in \overline{(X\setminus A)}$ so $x \in \overline{A}\cap\overline{(X\setminus A)}$ and $x \in\partial A$ so $x\in A^\circ\cup\partial A$
if $x \in A^\circ$ then $x \in A^\circ\cup \partial A$ so $\overline{A}\subset A^\circ\cup\partial A$
Best Answer
Based on the flaws suggested in the comments this I think (IMHO) this is an easier way to approach some parts of the proof.
To prove the line that $x \in ∂X \implies x \in \overline A $
Suppose $x$ is in the boundary of $A$ and $x$ is not in some closed set $B$ which contains $A$. Then $x \in B^c$ which is open and hence there is a neighbourhood $V_x$of $x$ which entirely avoids $A$ leading to a contradiction since every neighbourhood of $x$ must contains elements in $A$ and $A^c$.
Similar reasoning can be used to show that $x \in \overline A \implies x \in A^{\circ}$ or $x \in ∂X$.
Suppose $x \in \overline A$ and $x$ is an exterior point of $A$. Then there is a neighbourhood of $x$ which entirely avoids $A$. But then there is a closed set which contains $A$ but not $x$. This leads to a contradiction since $x \in \overline A \implies x$ is in every closed set containing $A$. Then $x$ is not an exterior point of $A \implies x$ is either an interior point or a boundary point of $A \implies x \in A^{\circ}$ or $x \in ∂X$