Definition:
Let $X\subseteq R$ and let $x'\in R$. We say that $x'$ is an adherent point of $X$ iff $\forall \epsilon >0, \exists x\in X$ s.t. $d(x′,x)\le \epsilon$. The closure of $X$ is denoted as $\overline X$ and is defined to be the set of all the adherent points of $X$.
Show that: $\overline{X} \cup \overline{Y} = \overline{X\cup Y}$
Proof: Assume that there exist a $z$ such that $z \in \overline{X} \cup \overline{Y} $ and $z \notin \overline{X\cup Y} $
since $z \notin \overline{X\cup Y} $, then $z$ is not an adherent point to $X\cup Y$
hence, $z$ is not adherent point to $X$ and to $Y$ in other words, $z \notin \overline{X}$ and $z \notin \overline{Y}$ but this contradicts with assumption
Hence $\overline{X} \cup \overline{Y} \subseteq \overline{X\cup Y}$
is my proof correct?