Let X be a topological space, and A sub set
how to prove that the following are equivalent:
1) A is connected
2) if A is decomposed into two open sets M and N such that $A=M\cup N$ and $$ (\overline{M}\cap N)\cup(M\cap \overline{N})=\emptyset $$ then $N=\emptyset$ or $M=\emptyset$
Thank you
On
On definitions: Following the link in the comments to the original definitions the OP is using (namely this French text) we see that $A$ is called a connected subset if it is connected in the subspace topology and a space is called connected whenever we write it as a disjoint union of two closed (or two open) subsets, one of them is empty.
Now suppose $A$ is connected, and write $A= M \cup N$ where $\overline{M}\cap N= M \cap \overline{N}=\emptyset$ (so called separated sets). Note that this already implies that $M$ and $N$ are disjoint.
Then $M$ is closed in $A$ (in its subspace topology), because if $x \in A$ lies in the closure of $M$ (in $A$), $x \in N$ is impossible (as $N$ misses $\overline{M}$) and so $x \in A=M \cup N$ implies $x \in M$. Similarly $N$ is closed in $A$ as well.
But then $M \cup N$ is (in $A$) just a decomposition of $A$ into two disjoint closed sets and so one of them is empty. Done.
The reverse is also clear, because if we write $A$ as a union of $C \cup D$, disjoint relatively open sets, $C$ and $D$ are separated in $X$ (if $x \in D$, $x \in U \cap A=D$ for some open subset $U$ of $X$, and this $U$ witnesses that $U \cap C = \emptyset$, so that $x \notin \overline{C}$; that $\overline{D} \cap C = \emptyset$ is symmetrical), and so one of them is empty by the assumption on separated sets.
A is disconnected iff exists open U,V with
not empty, disjoint U $\cap$ A, V $\cap$ A
and A = (U $\cup$ V) $\cap$ A. (1)
The disjoint sets are the M,N that decompose A.
Thus A is connected iff for all open U,V, (if
U $\cap$ A, V $\cap$ A are disjoint and A = (U $\cup$ V) $\cap$ A,
then U $\cap$ or V $\cap$ A is empty).
The conclusion follows be negating both sides of (1).