I'm practicing my proof-writing and was hoping you could let me know if this proof looks good. I would like to know if the proof is incorrect, if there are parts that are overly wordy/complicated, or if I'm missing some element of a proof that is helpful to see, if not strictly necessary.
The Prompt:
Are closures and interiors of connected sets always connected? (Look at subsets of $\mathbb{R^2}$.)
My Proof:
Let $A$ be a set. Suppose the closure of $A$, $\bar{A}$, is not connected, so that it is the union of two sets $B$ and $C$ s.t. $B$ and $C$ are disjoint. Since $\bar{A}$ is the closure of $A$, $\bar{A} = A \cup A_L$, where $A_L$ is the limit points of $A$. Therefore, $A \subset B \cup C$, and then let $B'$ be the elements of $B$ s.t. $B' \subset A$, and likewise let $C'$ be the set s.t. $C' \subset A$, so $A = B' \cup C'$. Then, since $B$ and $C$ are disjoint, $B'$ and $C'$ are disjoint, so $A$ is also not connected. Therefore since $\bar{A}$ being not connected implies $A$ is not connected, if $A$ is connected then $\bar{A}$ is connected. Thus, all closures of connected sets are connected.
Now, define a set $A$ s.t. $A = \{(x,y) \in \mathbb{R^2}|x \times y \ge{0}\}$. $A$ is a connected set. Although the origin is in $A$, it is not in the interior of $A$, $A^o$, as there is no neighborhood of $(0,0)$ s.t. all elements are a subset of $A$. Thus, $A^o = B \cup C$ where $B = \{(x,y) \in \mathbb{R^2}|x>0$ and $y>0\}$ and C = $\{(x,y) \in \mathbb{R^2}|x<0$ and $y<0\}$, where $B$ and $C$ are disjoint, so $A^o$ is not connected even though $A$ is. Therefore, interiors of connected sets are not always connected.