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**Plato** Suppose that $\displaystyle A$ is not connected. Then there is a separation of $\displaystyle A = H \cup K$.

Where $\displaystyle \begin{gathered} \emptyset \ne H \subseteq A\;\& \;\emptyset \ne K \subseteq A \hfill \\ \overline H \cap K = \emptyset \;\& \;\overline K \cap H = \emptyset \hfill \\

\end{gathered}$.

If $\displaystyle B \cap H = \emptyset $ then $\displaystyle H \cup \left( {B \cup K} \right)$ is a separation of $\displaystyle A\cup B$, contradiction.

So $\displaystyle \left( {B \cap H} \right) \cup \left( {B \cap K} \right) = B \cap A$ is also a separation of $\displaystyle A\cap B$.

Now you have to do the other.