Proving that the closure of a connected set in a metric space is connected. 
Proving that the closure of a connected set in a metric space is connected.

This was a proof that was done in class and I'm sure that it is wrong
Proof: Let $(X, d)$ be a metric space. Suppose that $Y \subseteq X$ is connected. We show that $\overline{Y}$ is also connected.
Suppose that $(\overline{Y}, d)$ was disconnected, then there exist open sets $U, V$ of $\overline{Y}$ such that $U \cup V = \overline{Y}$ and $U \cap V = \emptyset$ and $U \neq \emptyset$ and $V \neq \emptyset$.
Since $Y \subseteq \bar{Y}$ we thus have $U \cap Y$ and $V \cap Y$ to be clopen in $Y$ and $(U \cap Y) \cup (V \cap Y) = Y$. Assume that $U \cap Y \neq \emptyset$ then if $V \cap Y \neq \emptyset$ we have a contradiction.
So if $U \cap Y \neq \emptyset$ then we must have $V \cap Y = \emptyset$ and $U \cap Y = Y$. Since $V \cap Y =\emptyset$ we must have that $V$ contains only limit points of $Y$. 
But then for any $x \in V$ we have $B(x, r) \cap U \neq \emptyset$ for any $r > 0$. So any ball $B(x, r)$ contains points outside of $V$ contradicting the fact that $V$ is open in $(\bar{Y}, d)$. $\square$

I'm a bit confused by the last two lines. $V$ being open in $\bar{Y}$ means $V = \bar{Y} \cap A$ for some open $A$ in $(X, d)$, I don't see why any open ball $B(x, r)$ containing points outside of $V$ would contradict the fact that $V$ is open in $(\bar{Y}, d)$ (It sure would contradict the fact that $V$ is open in $(X, d)$ though)
For example take $(\mathbb{R}, d)$ with the usual metric, and consider $\bar{Y} = [0, 1]$, and the open set $V = [0, \frac{1}{2})$ in $\bar{Y}$ then $B(0, r) = (-r, r)$ contains points outside of $V$ so by the above logic $V$ cannot be open in $\bar{Y}$ (which is wrong because $V$ is certainly open in $\bar{Y}$)
Am I correct in saying that there is an error in this proof?
 A: The proof you quote makes it needlessly complicated: we know $U \cap Y = Y$, $Y \cap V = \emptyset$. 
continuing: $V$ being open in $\overline{Y}$ means that there exists $V'$ open in $X$ with $V' \cap \overline{Y} = V$. 
Note that $Y \cap V =\emptyset$ implies $Y \subseteq X\setminus V'$ (otherwise for some $y$, $y \in Y \cap V' \subseteq \overline{Y} \cap V' = V$, which is not the case), and so $\overline{Y} \subseteq X\setminus V'$, (as the latter set is closed) and so $V' \cap \overline{Y} = V = \emptyset$. Contradiction and done. 
No need to mention balls at all; this works in any topological space.
A: 
The proof is valid but writing $B(x, r) \cap U \neq \emptyset$ is a
  bit confusing. To clarify the argument, I would replace the last
  paragraph of the proof with this:

But then for any $x \in V$ we have $B(x, r) \cap Y \neq \emptyset$ for any $r > 0$. So any ball $B(x, r)$ contains points outside of $V$ since $Y \subset U$ and  $U \cap V = \emptyset$. But this is absurd since $V$ is open in $(\bar{Y}, d)$.
We also get a contradiction if we assume that $V \cap Y \neq \emptyset$, so it is not possible to disconnect $\bar{Y}$ with two open sets $U$ and $V$.
 $\square$
Note: We make use of the fact that any point $w$ in an open set $W$ of a metric space belongs to an open ball $B(w, r)$ that is wholly contained in $W$.


So to show $\bar{Y}$ is connected the above examines opens sets. We
  can argue by focusing on just the closure
  operator used in topology.

In what follows $X$ is assumed to be a disconnected space with closed sets $K$ and $L$ providing a non-trivial decomposition.
Now any subspace of $T$ of $X$ with points in both $K$ and $L$ is manifestly disconnected along with its closure $\bar{T}$. 
If $S$ is a subspace of $X$ that is contained in, say  $K$, then $\bar{S} \subset K$.
Suppose we can write $X = \bar{Y}$. Then $Y$ can't be connected. If it was connected it would have to be wholly contained in either $K$ or $L$. In one case we would have that $X = K$ and in other case $X = L\,$. But either one of these identities contradicts the assumption that we started with a non-trivial decomposition of $X$. 
