Direct proof of $A \subseteq \overline{A}$ I proved $A \subseteq \overline{A}$ by showing $\overline{A}^c \subseteq A^c$: Let $x \in \overline{A}^c $ then $x$ is not an accumulation point of $A$ therefore there is an open nbhood $N$ of $x$ with $N \cap A = \varnothing$ or equivalently: $N \subseteq A^c$.
This is fairly short. I tried do it directly but failed. Is it possible to show $A \subseteq \overline{A}$ directly? (Assume $x \in A$ and $N$ an open nbhood of $x$. Then $N$ must contains a point of $A$)
The definition of closure is the set and all its accumulation points.
 A: If you've defined $\overline A$ to be the set $A$ together with all its accumulation points, then by definition, $A$ is a subset of $\overline A.$
If instead you've defined it to be the set of all $x$ in the overlying space such that every neighborhood of $x$ intersects $A,$ then note that for any $x\in A$ and any neighborhood $N$ of $x$ we have $x\in N,$ so $A\cap N\neq\emptyset,$ and so $x\in \overline A.$
It's a good exercise to show that these two definitions are equivalent. (Hint: $x$ is an accumulation point of $A$ if and only if every neighborhood of $x$ intersects $A$ at some point different from $x$.)
A: It depends on what specific definition you are taking for $\overline A$.  For example, the definition I would take is that $\overline A$ is the intersection of all the closed sets that contain $A$.  Then as each of the sets you are intersecting contains $A$ we can conclude that their intersection $\overline A$ also contains $A$.
A: Let $A'$ denote the set of accumulation points of $A$. Then you've defined $$\bar A=A\cup A'$$
This means $\bar A\supseteq A$ at once.
