I have to prove that if $E$ is a connected space, then so is $\overline{E}$ (the closure of $E$) a connected space. I tried to prove the contrapositive. So suppose that $\overline{E}$ is not connected in a metric space $X$. This implies there are $A\subset X$ and $B\subset X$ such that $A\cap B=\emptyset$, $A\cap \overline{E}\neq \emptyset$, $B\cap \overline{E}\neq \emptyset$ and $\overline{E}\subset A\cup B$. I will try to prove that this $A$ and $B$ will also work for $E$. Suppose that $A\cap E=\emptyset$ and $B\cap E=\emptyset$, this implies that $A\cup B$ is a subset of the limit points of $E$, because $A\cap \overline{E}\neq \emptyset$ and $B\cap \overline{E}\neq \emptyset$. This contradicts the fact that $\overline{E}\subset A\cup B$, so we have to conclude that $A\cap E\neq \emptyset\neq B\cap E$. Since $E\subset\overline{E}\subset A\cup B$, we have that $E$ is also not connected.
My question is whether my proof is correct, because I have a little bit doubt. If it's not correct, what is the best I can do? Thanks in advance!
Edit1: The definition we use for a connected space $E$ in a metric space $X$ is that if $E$ is connected, then there are no open $A,B\subset X$ such that $A\cap B=\emptyset$, $A\cap E\neq \emptyset\neq B\cap E$ and $E\subset A\cup B$.
Edit2: $A$ and $B$ must be open.
Best Answer
A subset $E⊆X$ is connected iff the only continuous functions $f:E→\{0,1\}$ are the constant ones. A continuous function $g:X→Y$ where $Y$ is Hausdorff is uniquely determined by a dense subset of $X$, and $\{0,1\}$ with the discrete metric is evidently Hausdorff. $E$ is dense in $\overline E$ and continuous functions $f:E→\{0,1\}$ are constant by assumption.
Another proof my proceed by contraposition. Suppose that $\overline E$ is disconencted by $U,V$. Then $E$ being dense has nonempty intersection with both $U,V$, so it is disconnected by $U'=E\cap U\;,\; V'=V\cap E$, for $U',V'$ are disjoint nonempty and relatively open in $E$, and $E=V'\cup U'$ by $\bar E=V\cup U$.