Every compact Hausdorff space can be expressed as a disjoint union of finitely many open sets. Let $X$ be a compact Hausdorff space. 
Can we express as disjoint union of open sets ?
I got a proof for that.
Since $X$ is compact there exist a  finite sub collection of open sets $\{W_{i}\}_{i=1}^{n}$ that covers $X$.  Construct disjoint open sets $W_{i}'$ for $i=1, \cdots,n$ as follows :
\begin{eqnarray*}
W_{1}'= & W_{1} \\
W_{k}'= & W_{k}-\displaystyle \bigcup_{j=1}^{k-1}\overline{W_{j}'}'
\end{eqnarray*}
Then by the construction $W_{i}'$  are disjoint. 
Suppose $x \in W_{i}$ only then $x$ belongs to $W_{i}'$. Now let $x$ be an element of more than one $W_{i}$. Let $i$ be the least index for which $x \in W_{i}$ then  $x $ belongs to that $W_{i}'$  and does not belongs to any other $W_{j}'$. Thus the collection $\{ W_{i}' \}_{i=1}^{n}$ covers $X$.
Is this proof correct ? If it is other questions are immaterial. I know that I have not used Hausdorff condition in the proof.
Can we have a counter example ?
Can we apply some conditions on compact Hausdorff space so that it can be expressed as disjoint union of open sets ?

You can always do it with $n=1$ and $W_1=X$: if $X$ is connected, this is the one and only possibility. Otherwise, your procedure fails in general, your error lying in the fact that $\{W_i',\:\, i\in\Bbb N\}$ may not be a covering. Also, the entire idea of a non-trivial case $n\ge2$ fails catastrophically: again, see the concept of connected space, and related concepts such as connected components of a topological space.
