$A\cup B$ is connected proof

Question

Let $A$ and $B$ be connected subspaces of a topological space $(X,\tau)$. If $A\cap B\neq\emptyset$, prove that the subspace $A\cup B$ is connected.

If the subspace $A\cup B$ is not connected, then there exist, $\mathscr{U},\mathscr{V}\subset X$ such that $\mathscr{U}\cup\mathscr{V}=A\cup B$ and $\mathscr{U}\cap\mathscr{V}=\emptyset$. $\mathscr{U},\mathscr{V}$ must belong either to $A$ or $B$, like $\mathscr{U}\in A$, which contradicts the fact $A$ and $B$ are connected. Therefore $A\cup B$ is connected.

Questions:

Is my proof right? If not. How should I prove the statement?

Thanks in advance!

Answer 1

A subspace $Y$ of a topological space is disconnected (w.r.t. subspace topology) if there exist two non-empty open sets $U, V$ in the subspace topology of $Y$ such that $U\bigcup V=Y$ and $U\bigcap V=\varnothing$. A representation of $Y$ as a union of two non-empty sets which are open in subspace topology is called a disconnection.

Here $Y=A\bigcup B$ with $A\bigcap B\not=\phi$ and $A,B$ are connected subspaces of $X$. So if possible, let $Y$ be disconnected w.r.t. subspace topology, then we can find two sets $U, V$ having the properties of 1st paragraph. Now since $A$ is connected, $A $ is contained in one of the open sets, say $U$ (otherwise $A=(A\cap U)\bigcup (A\cap V)$ will be a disconnection of $A$). Similarly, $B$ is also contained in one of the sets $U, V$. Now since $A\bigcap B\not=\varnothing$ we can say $B$ is contained in $U$. Hence $V=\varnothing$. Therefore $A\bigcup B$ is connected.