$x \in \overline{E}$ iff $B_r(x) \cap E \neq \emptyset$. Question about proof. Proposition: $x \in \overline{E}$ iff $B_r(x) \cap E \neq \emptyset$
Proof: $x \in \overline{E}$ iff $x \in E \cup E'$.
In other words, $x$ has to be in either $E$ or $E'$.
If $x \in E$, then $B_r(x) \cap E \neq \emptyset$, for any $r>0$ since $x \in E$.
If $x \in E'$, then $B_r(x) \cap E -\{x\} \neq \emptyset$, for any $r>0$ by definition of a limit point.
The proof concludes by concluding that $B_r(x) \cap E \neq \emptyset$ for all $r>0$.
My question is: Why do we take the less restrictive statement to be true? 
 A: Proposition. $x\in\overline{E}$ if and only if, for all $r>0$, $B_r(x)\cap E\ne\emptyset$.
According to your question, you define $\overline{E}=E\cup E'$, where $E'$ is the set of limit points of $E$.
Proof. $(\Rightarrow)$ If $x\in E$ we have $x\in B_r(x)\cap E$, because $x\in B_r(x)$, for any $r>0$. Suppose $x\in E'$; then, given $r>0$, $B_r(x)\cap E\ne\emptyset$ by definition of limit point.
$(\Leftarrow)$ If $x\in E$ there's nothing to prove, so we can assume that $x\notin E$. Then $x$ satisfies the definition of limit point of $E$; therefore $x\in E'$ and so $x\in\overline{E}$.

Note that the result is false if you interpret the given statement as “$x\in\overline{E}$ if and only if, for some $r>0$, $B_r(x)\cap E\ne\emptyset$”. For instance, in the reals, $2$ satisfies the condition when $E=(0,1)$ and we take $r=5$. But of course $2\notin\overline{E}$.
A: We can actually prove a more general statement ($B_r(x)$ presupposes we have a metric, but a similar statement is true in all topological spaces)
Here the definiton of $\overline E$ used is: the intersection of all closed sets containing $E$. This should coincide with your definition (a good exercise for you).
Claim: Let $A$ be a subset of the topological space $X$. Then $x \in \overline A$ if and only if every open set $U$ containing $x$ intersects $A$.
Proof: Suppose that $x \in \overline A$. Suppose, by way of contradiction, that there exists an open set $U \ni x$ where $U \cap A = \emptyset$. This means that $A \subseteq X \setminus U$ which implies that $\overline A \subseteq \overline {X \setminus U} = X \setminus U$ (the latter equality is true because $U$ is open implies $X \setminus U$ is closed, and the closure of a closed set is itself). But since $x \in U$, it follows that $x \notin \overline A$, a contradiction. conclude that every open set $U$ containing $x$ intersects $A$.
Conversely, suppose that every open set $U$ containing $x$ intersects $A$. Suppose, by way of contradiction, that $x \notin \overline A$. Then $U = X \setminus \overline A$ is open, and $x \in U$. By hypothesis, $U \cap A \neq \emptyset$, but this is a contradiction to our construction of $U$ not containing $\overline A$. Conclude that $x \in \overline A$.
