The closure equals the set together with its limit points, equals the set of points every neighbourhood of which meets it, and is the smallest closed superset; a set is closed iff it contains its limit points

$A \subseteq E$: for $x \in A$ and any $\varepsilon > 0$ one has $x \in N_\varepsilon(x) \cap A$, so that intersection is nonempty.

Let $x \in \mathbb{R} \setminus E$; by the definition of $E$ there is a real $\varepsilon > 0$ with $N_\varepsilon(x) \cap A = \varnothing$.

Let $F$ be closed with $A \subseteq F$, and let $x \in \mathbb{R} \setminus F$; since $\mathbb{R} \setminus F$ is open there is a real $\eta > 0$ with $N_\eta(x) \subseteq \mathbb{R} \setminus F$.

For every $y \in N_\varepsilon(x)$ the radius $\delta := \varepsilon - |y - x|$ is positive and $N_\delta(y) \subseteq N_\varepsilon(x)$, so $N_\delta(y) \cap A = \varnothing$ and $y \notin E$; hence $N_\varepsilon(x) \subseteq \mathbb{R} \setminus E$, and since $x$ was an arbitrary point of $\mathbb{R} \setminus E$ that set is open, that is, $E$ is closed.

From $N_\eta(x) \subseteq \mathbb{R} \setminus F \subseteq \mathbb{R} \setminus A$ we get $N_\eta(x) \cap A = \varnothing$, so $x \notin E$; hence $\mathbb{R} \setminus F \subseteq \mathbb{R} \setminus E$, that is, $E \subseteq F$, for every closed $F \supseteq A$.

By steps 1.1 and 2.1 the set $E$ is a closed superset of $A$, so $\overline{A} \subseteq E$ by the leastness in [L3]; and $\overline{A}$ is itself a closed superset of $A$ by [L3], so step 2.2 applied to $F = \overline{A}$ gives $E \subseteq \overline{A}$. Hence $\overline{A} = E$, which is claim 1.

$E = A \cup A'$: if $x \in E$ and $x \notin A$ then for every $\varepsilon > 0$ some $a \in N_\varepsilon(x) \cap A$ exists, and $a \ne x$ because $x \notin A$, so $a \in N^{*}_\varepsilon(x) \cap A$ and $x \in A'$; conversely $A \subseteq E$ by step 1.1, and $A' \subseteq E$ because $N^{*}_\varepsilon(x) \subseteq N_\varepsilon(x)$. Combining with step 3.1 gives $\overline{A} = A \cup A'$, which is claim 2.

Claim 4: if $A$ is closed then $A$ is a closed superset of itself, so $\overline{A} \subseteq A$ by [L3], while $A \subseteq \overline{A}$ by [L3], whence $A = \overline{A}$; conversely if $A = \overline{A}$ then $A$ is closed because $\overline{A}$ is. Finally $A = \overline{A}$ says $A = A \cup A'$ by step 4.1, and $A = A \cup A'$ holds exactly when $A' \subseteq A$.

Claim 3 is [L3] restated, and claims 1, 2 and 4 are steps 3.1, 4.1 and 5.1, so all four hold.