Let $X$ and $Y$ be arbitrary subsets of $\textbf{R}$. Then
(a) $X\subseteq\overline{X}$
(b) $\overline{X\cup Y} = \overline{X}\cup\overline{Y}$
(c) $\overline{X\cap Y}\subseteq\overline{X}\cap\overline{Y}$
MY ATTEMPTS
(a) The set $\overline{X}$ denotes the set of all adherent points of $X$. In particular, it contains the elements of $X$.
This is because, no matter which $\varepsilon > 0$ one chooses, if $x\in X$, there is an element of $X$ (namely, $x$ itself) such that $|x - x| = 0 < \varepsilon$. Thus $X\subseteq\overline{X}$.
(b) Let us prove the inclusion $(\subseteq)$ first.
If $a\in\overline{X\cup Y}$, no matter how small one chooses $\varepsilon > 0$, there is a number $b\in X\cup Y$ such that $|a - b| \leq \varepsilon$.
If $b\in X$, then $a\in\overline{X}$. If $b\in Y$, then $a\in\overline{Y}$. In both cases, $a\in\overline{X}\cup\overline{Y}$, and we are done.
Let us now prove the inclusion $(\supseteq)$
If $a\in\overline{X}\cup\overline{Y}$, then either $a\in\overline{X}$ or $a\in\overline{Y}$. Let us assume $a\in\overline{X}$.
Thus, no matter which $\varepsilon > 0$ one chooses, there is an element $b\in X\subseteq X\cup Y$ such that $|a-b|\leq\varepsilon$. Hence $a\in\overline{X\cup Y}$.
Analogously, let us suppose that $a\in\overline{Y}$. The same reasoning shows that $a\in\overline{X\cup Y}$. Therefore $\overline{X}\cup\overline{Y}\subseteq\overline{X\cup Y}$.
(c) Finally, let us prove the last assertion.
Let us suppose that $a\in\overline{X\cap Y}$. Then no matter which $\varepsilon > 0$ one chooses, there exists an element $b\in X\cap Y$ such that $|a - b|\leq\varepsilon$. Since $b\in X$ and $b\in Y$, this means that $a\in\overline{X}$ and $a\in\overline{Y}$, that is to say, $a\in\overline{X}\cap\overline{Y}$, just as desired.
I am little bit new to this. Could someone please verify if I am reasoning right?
Note that you can prove (b) and (c) just by using (a)
both $X$ and $Y$ are a subsets of $X\cup Y$. from that you have that $\overline{X}$ , $\overline{Y} \subseteq \overline{X\cup Y}$. follows that their union is a subset of $\overline{X\cup Y}$ as well. on the other hand, $X\cup Y \subset \overline{X} \cup \overline{Y} $ since $X \subset \overline{X}$ and $Y \subset \overline{Y}$.
use the same reasoning for c