Let $\{X,\tau\}$ be a topological space, $E\subset X$; prove that $cl(E)$ is connected if and only if $E$ is not union of two non-empty subsets $A$ and $B$ such that $cl(A)$∩$cl(B)$= $\emptyset$
when i prove it i get that at begining $E\subset cl(E)\subset cl(E)$ so i prove it by contradiction but by doing some steps i get that $cl(E)\subset cl(E)\subset cl(E)$ this is not correct because i'm around in the same point
So what i can do to avoid that mistake i mean what to do at the begining ?
Source:Topology without tears\Sidney A. Morris.
To make some subscripts easier to read, let $cl_X(E)=F.$
(1). If $E=A\cup B$ where $A, B$ are non-empty and $cl_X(A)\cap cl_X(B)=\phi$ then (since, also, $A$ and $B$ are subsets of $F$) we have $$F=F\cap cl_X(E)=F\cap (cl_X(A\cup B))=$$ $$=F\cap (cl_X(A)\cup cl_X(B))=$$ $$=(F\cap cl_X(A))\cup (F\cap cl_X(B))=$$ $$=cl_F(A)\cup cl_F(B)$$ so the $space$ $F$ is the union of two disjoint non-empty sets ($cl_F(A)$ and $cl_F(B)$) that are closed in $F$ so $F$ is disconnected.
(2). If $F$ is disconnected then $F=C\cup D$ for some non-empty disjoint $C,D\subset F$ such that $C=cl_F(C)$ and $D= cl_F(D).$
Let $A=C\cap E$ and $B=D\cap E.$ We have $A\cup B=(C\cap E)\cup (D\cap E)=(C\cup D)\cap E=F\cap E=E.$
Now $A=\phi\implies B=E\implies D=cl_X(D)\supset cl_X(B)=cl_X(E)=F$ which implies $C=\phi,$ contrary to hypothesis. So $A\ne \phi.$ Similarly, $B\ne \phi.$
Since $A,B$ are subsets of $F$ we have $cl_X(A)\subset cl_X(F)=cl_X(cl_X(E))=cl_X(E)=F$ and $cl_X(B)\subset cl_X(F)=cl_X(cl_X(E))=cl_X(E)=F.$
Therefore ( since also $A\subset C$ and $B\subset D$) we have $$cl_X(A)\cap cl_X(B)=$$ $$=(F\cap cl_X(A))\cap (F\cap cl_X(B))=$$ $$=cl_F(A)\cap cl_F(B)\subseteq$$ $$\subseteq cl_F(C)\cap cl_F(D)=\phi.$$