Before I start, let me give three definitions.
Definition (Separation): Let $X$ be a topological space. A separation of $X$ is a pair $U, V$ of disjoint non-empty open subsets of $X$ whose union is $X$. We will write a separation $U, V$ as $(U, V)$.
Definition: A topological space $X$ is said to be connected if there does not exist a separation of $X$.
Definition: A subset of a topological space $X$ is said to be connected if it is connected in the subspace topology.
My question revolves around this statement which I think is true.
Claim: Let $X$ be a topological space; and let $Y$ be a subspace of $X$. Suppose $(A, B)$ is a separation of $X$. Then $(A \cap Y, B \cap Y)$ is a separation of $Y$ if and only if both $A \cap Y$ and $B \cap Y$ are non-empty.
This is a claim that sparked in my head when proving a theorem from Munkres' book, which is why I'm doubting its truth (as I'm very handicapped at topology). I think it may be true so I tried to prove it and the logic seems fine to me.
Proof. ($\implies$). This direction is trivial as suppose $(A \cap Y, B \cap Y)$ is a separation of $Y$, then by definition $A \cap Y$ and $B \cap Y$ are non-empty.
($\impliedby$). Suppose $A\cap Y$ and $B \cap Y$ are non-empty. Since $(A, B)$ is a separation of $X$, $A$ and $B$ are open in $X$. Therefore, the two sets $A \cap Y$ and $B \cap Y$ are relatively open in $Y$. Moreover, observe that the two sets $A \cap Y$ and $B \cap Y$ are disjoint and their union is $Y$. Therefore, $(A \cap Y, B \cap Y)$ forms a separation of $Y$. $\rule{0.5em}{0.5em}$
My question is: Would there be any counterexamples to the claim as I could not think about one. Moreover, if there is a counterexample, would anyone be kind enough to reveal which part of the proof that is flawed.
If anyone has a thought on strengthening the conditions of this claim, I would also welcome that.
The claim is true; note indeed that $A \cap Y$ and $B \cap Y$ are relatively open in $Y$, still disjoint, and if $A \cup B $ covered $X$, their intersections with $Y$ cover $Y$; so the only "concern", or possible problem, with being a separation is that both parts should be non-empty. So your proof is correct, and if both parts are non-empty, we get a separation of the subspace.
A consequence of this observation is that if $X$ disconnected and $(U,V)$ is a disconnection, then for every connected subspace $Y$ of $X$, either $Y \subseteq U$ or $Y \subseteq V$. This is often used in proofs involving connectedness.