Prove that the closure of a connected space is connected (Topology) Let A be a connected subspace of a topological space (X,T), I start by assuming that cl(A) is disconnected.  Therefore there exists two open sets of T (call them U and V) such that U ∩ V = ∅ and U ∪ V = cl(A).
If U and V are in (A, T_A), then they're connected -- a contradiction.
If U and V are in (A', T_A'), then U ∩ V may = ∅, but U ∪ V ≠ cl(A) -- contradiction.
If U is in (A, T_A) and V is in (A', T_A'), then U ∩ V = ∅ and U ∪ V = cl(A) iff U = A and V = A'.  But this implies that A ∩ A' = ∅ and A ∪ A' = cl(A).
The only way for the cl(A) to be connected is for A ∩ A' = ∅ to be a false statement, but I cannot find a way to show that A ∩ A' ≠ ∅ if A is a connected subspace.
Anyone have any ideas?
 A: Here is a useful characterisation of connectedness for your problem.

Theorem: The space $X$ is disconnected if and only if there is a continuous surjection $X \to \{0,1\}$ where the codomain carries the discrete topology.

Now suppose $A^-$ is disconnected but $A$ is connected. That means there is a continuous surjection $f \colon A^- \to \{0,1\}$. Of course then the restriction $f \colon A \to \{0,1\}$ is also continuous. But since $A$ is connected the restriction cannot be surjective. So without loss of generality $f(A) = \{0\}$. But here is a characterisation of continuity.

Theorem: The function $g \colon X \to Y$ is continuous if and only if $g(B^-) \subset g(B)^-$ for every subset $B \subset X$.

So we must have $f(A^-) = \{0\}^- = \{0\}$. But this contradicts how $f$ is surjective.
A: We will restate (with the same intention but some 'smoothing out') the OP's logic/proof.
If $cl(A)$ of a subspace $A$ of a topological space $X$ is disconnected, then $A \subset cl(A)$ must also be disconnected.
Since $cl(A)$ is disconnected, we can find two open sets $U$ and $V$ in $X$ satisfying the following:
$\tag 1 U \cap V \cap cl(A) = ∅$
$\tag 2  (U \cup V) \cap cl(A) = cl(A)$
$\tag 3 U \cap cl(A) \ne \emptyset$
$\tag 4 V \cap cl(A) \ne \emptyset$
We claim the open sets $U$ and $V$ also 'disconnect' $A$. To show this we have to verify that $\text{(1) thru (4)}$ hold when $cl(A)$ is replaced with $A$. It is easy to see that both $\text{(1)}$ and $\text{(2)}$, reformulated, are true.
To show that both $U \cap A$ and $V \cap A$ are nonempty sets, assume that, say, $V \cap A = \emptyset$. But by $\text{(4)}$ there is a point $v \in V$ that is also in $cl(A)$, with $V$ an open set containing $v$ that doesn't intersect $A$. This is absurd, and we can also argue in the same way that $U \cap A \ne \emptyset$.
So both $\text{(3) thru (4)}$ are also true for the subspace $A$.
A: Suppose $U,V$ are open sets such that $U\cap V=\emptyset$ and $cl(A)\subseteq U\cup V$. We would like to show that either $U\cap cl(A)=\emptyset$ or $V\cap cl(A)=\emptyset$.
Note that, since $A$ is connected, we may assume without loss of generality that $U\cap A=\emptyset$.
If $U\cap cl(A)\neq \emptyset$, there is an element $a\in U\cap cl(A)$. Notice however that in this case $U$ would be a neighborhood of $a$ that does not contain any element of $A$, so $a\not\in cl(A)$. A contradiction.
Therefore, $U\cap cl(A)=\emptyset$. This shows that $cl(A)$ is connected.
A: 
Following the crystal clear title of this post, the OP's provides some
proof details that appear to be mangled logic.
This is what I would like to replace the question/post with:

Let $A$ be a connected subset of $X$. I want to show that if $U$ and $V$ are two open sets in $X$ such that
$\tag 1 cl(A) = U \cap cl(A) \; \bigcup \; V \cap cl(A)$
and
$\tag 2 U \cap cl(A) \; \bigcap \; V \cap cl(A) = \emptyset$
then either $cl(A) \subset U$ or  $cl(A) \subset V$.
I have to use the fact that $A$ is connected, but I'm running into problems trying to put this argument together.

Answer
Rather than answer the question directly, we will put together some general theory that can be used, in a straightforward manner, to prove that the closure of any connected space is also connected.
If $X$ is any non-connected topological space, then two nonempty clopen sets $X_1$ and $X_2$ can be found that partition $X$. Now if $A$ is any dense subspace of $X$, it can't be wholly contained in either $X_1$ or $X_2$. Indeed, if $A$ was contained in, say $X_1$, the closure of $A$ would also be contained in that set ($X_1$ is closed). But then $A$ can't be a dense subset of $X$. So we can state the following:
Proposition: If $X$ is any non-connected topological space, then every dense subspace of $X$ must also be disconnected.
