Problem: Suppose $A \subset M$ is connected where $M$ is a metric space. Show that $A, \emptyset $ are the only subsets of $A$ that are clopen relative to $A$.
My thoughts:
The empty set is always open and closed. Since $M$ is closed and open, and $A=A \cap M$, then $A$ is open relative to $A$ (and similarly closed).
Now supposing there was another clopen set $U \neq \emptyset, A$, Then I am imagining I need to use $U$ and $U^c$ to show that there is a separation of $A$. My definition of a separation is that $A = U \cup U^C$ where $U \cap closure(U^c) = \emptyset$ and $closure(U) \cap U^c = \emptyset$ which would contradict that $A$ is connected.
If $U$ is clopen relative to $A$, then there exists $V$ open in $M$ with $U=V \cap A$, and $W$ closed with $U=W\cap A$, how can I show $U \cap closure(U^c) = \emptyset$ and $closure(U) \cap U^c = \emptyset$ ?
Edit: Worth noting this problem has been asked many times but not with relative topology :)
Note: I'm going to use a bar over a set to represent closure. (Thus the closure of $U$ is $\overline U$.) Also, complements will always be relative to $A$, but open/closedness relative to $M$ (matching the usage in your question).
$U^c = V^c \cap A$, so $U^c \subset V^c$, thus $\overline {U^c} \subset \overline {V^c} =V^c$ (since $V^c$ is closed). Thus $\overline {U^c} \cap V = \varnothing$, so $\overline {U^c} \cap U = \overline {U^c} \cap V \cap A = \varnothing$.
Similarly, since $U \subset W$, $\overline {U} \subset \overline {W} =W$, so $\overline U \cap W^c = \varnothing$. $U^c = W^c \cap A$, thus $\overline {U} \cap U^c = \overline {U} \cap W^c \cap A = \varnothing$.