Show that $A \subset B \implies \overline{A} \subset \overline{B}$

Question

NOTE: I know that a question asking for help to prove this same property already exists, but I would like an answer specifically based on the definition(s) and / or remark below, please.

Definition 1: A point $x \in \mathbb{R}$ is a point of closure of a set $E \subset \mathbb{R}$ if
$\quad \forall \ \delta>0,\; \ \exists \ y \in E \ \;$ s.t. $ \ |x-y| < \delta$.
Equivalently, $x$ is a point of closure of $E$ if every open interval containing $x$ also contains a point of $E.$
We call the set of all points of E the closure of $E$ and denote it by $\overline{E}.$

Remark: Every point in $E$ belongs to its closure. Particularly, $E \subset \overline{E}$.

Definition 2: $E$ is closed if $E=\overline{E}$.

Question: Show that $A \subset B \implies \overline{A} \subset \overline{B}$.

Attempt: $A \subset B \implies A \subset B \subset \overline{B} \implies A \subset \overline{B}$. If $A$ is closed then $A= \overline{A} \implies \overline{A} \subset \overline{B}.$

So I can do this when $A$ is closed but I'm not certain on how to use either of the definitions / remark to show that it holds when $A$ is open.

Answer 1

Let $p$ belong to the closure of $A$. Then every interval $I$ including $p$ has some $a \in A \cap I.$ But then $a \in B \cap I.$ That makes $p$ a closure point for $B$, by your definitions. (We did not use definition 2 here.)

Answer 2

Let $x \in \overline{A}$. Then $x \in A \cup A'$ where $A'$ is the set of limit points of A. If $x\in A \subset B$, then $x \in B \subset \overline B$. So let $x \in A'$. Then for any $\epsilon$ nbd of $x$, there exists $y \neq x$ such that $y \in A$. But since $A \subset B$, $y \in B$. Hence, $x \in B' \subset \overline B$. This completes the proof.