This question might have been answered before but I need different proof using the following fact:

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> Suppose $C$ is a connected set in a metric space (M,d). Let $A$ and $B$ be two separated sets in $(M,d)$, i.e., $A \cap \bar{B} = \emptyset = A \cap \bar{B} $ such that $C \subseteq A \cup B$. Show that $C \subseteq A$ or $C \subseteq B$.
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We want to show if $C$ is a connected set then closure of $C$, i.e. $\bar{C}$ is connected.

Claim : $\bar{C}$ is connected

To show the claim we assume that $\bar{C}$ is disconnected. If we do so, we need to reach a contradiction which is possibly would be $C$ is disconnected.

Proof:

Suppose $C$ is a connected set and $\bar{C}$ is not connected.

Since $\bar{C}$ is a disconnected set, there exist two non-empty sets $A$ and $B$ which satisfy the following:

1- $A \ne \emptyset$ and $B \ne \emptyset$

2- $A \cap \bar{B} = \emptyset = A \cap \bar{B} $ 

3- $\bar{C}= A \cup B$

Using the fact $C \subseteq \bar{C}$ $\rightarrow C \subseteq A \cup B$.

Now using the fact that $A$ and $B$ are two separated sets and $ C \subseteq A \cup B$, we can use the very first fact so

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> $C \subseteq A$ or $C \subseteq B$
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WLOG, asume that $C \subseteq A$ so $C \cap B = \emptyset $

From this point how can we prove $C$ is disconnected to reach to contradiction to the first assumption.

My guess is we need two non-empty sets $U$ and $V$ which have the following criteria:

1- $U \ne \emptyset$ and $V \ne \emptyset$

2- $U \cap \bar{V} = \emptyset = U \cap \bar{V} $ 

3- $C= U \cup V$

But what are they?