Engelking's book- proof of The Cantor-manifold theorem (1.9.8.) In the proof of the theorem mentioned in the title, from "Dimension Theory" Engelking's book, I don't understand the following claim:
$X$ is an $n-$dimensional Cantor manifold if and only if $\mathrm{ind}(X)\leq n$ and whenever $X=X_1\cup X_2$, where $X_1$ and $X_2$ are proper closed subsets of $X$, then $\mathrm{ind}(X_1\cap X_2)\geq n-1$.
Recall: A compact metric space such that $\mathrm{ind}(X)=n\geq 1$ is an $n-$ dimensional Cantor-manifold if no closed subset $L$ of $X$ satisfying the inequality $\mathrm{ind}(L)\leq n-2$ separates the space $X$, i.e., if for every such set, $X\setminus L$ is connected.
I'd appreciate any help! Thanks.
 A: Suppoose $X$ is an $n$-dimensional Cantor manifold and $X=X_1\cup X_2$ where $X_1$ and $X_2$ are proper closed subsets of $X$.  Then $X\setminus(X_1\cap X_2)$ is disconnected, since it can be partitioned into the closed subsets $X_1\setminus X_2$ and $X_2\setminus X_1$.  Thus $\operatorname{ind}(X_1\cap X_2)$ cannot be less than or equal to $n-2$, so it must be at least $n-1$.
Conversely, suppose $X$ is a compact metric space such that $\mathrm{ind}(X)\leq n$ and whenever $X=X_1\cup X_2$, where $X_1$ and $X_2$ are proper closed subsets of $X$, then $\mathrm{ind}(X_1\cap X_2)\geq n-1$.  Suppose a closed subset $L\subseteq X$ separates $X$; partition $X\setminus L$ into nonempty closed subsets $A$ and $B$ and let $X_1=L\cup A$ and $X_2=L\cup B$.  Then $X_1$ and $X_2$ are proper closed subsets of $X$ and $X=X_1\cup X_2$, so $\operatorname{ind}(X_1\cap X_2)\geq n-1$.  That is, $\operatorname{ind}(L)$ is not less than or equal to $n-2$.
To see that $\mathrm{ind}(X)= n$ it suffices to show that $\mathrm{ind}(X)\geq n$.  If $\mathrm{ind}(X)\leq n-1$, then for any distinct points $x,y\in X$, there is a partition $L$ between them with $\operatorname{ind}(L)\leq n-2$, so that $X\setminus L$ partitioned into closed sets $A$ and $B$ which contain $x$ and $y$ respectively.  Then $X_1=L\cup A$ and $X_2=L\cup B$ are proper closed subsets of $X$ with $X_1\cup X_2=X$, and so $\operatorname{ind}(L)=\operatorname{ind}(X_1\cap X_2)\geq n-1$.  This is a contradiction.
(Note that the last argument requires $X$ to have at least two points; indeed, the reverse implication is incorrect if $X$ is a singleton or empty.)
