Noncompact topological space such that its Alexandroff compactification is not connected Example of a noncompact topological space X such that its Alexandroff compactification X* is not connected.
I proved that it X* is connected, than X is noncompact. I need a counterexample for the other direction. 
 A: If $X$ is connected and non-compact then, as $X$ is dense in $X^\ast$, $X^\ast$ is also connected. For compact $X$, $X^\ast$ is not connected, as $\infty$ is an isolated point, and $\{\infty\}$ is a non-trivial clopen set. 
If $X=\Bbb N$ (non-compact) then in $X^\ast$, $\{0\}$ is still a non-trivial clopen set in $X^\ast$, so $X^\ast$ is not connected. (it's a convergent sequence). 
More generally, if $X$ has a non-empty compact clopen subset then $X^\ast$ will not be connected. But $X$ can be disconnected, like $X=(0,1) \cup (2,3)$ and $X^\ast$ can then still be connected (two touching circles in this case). So $X^\ast$ connected does not say much about $X$ being connected: it could be or it could not be. We only know that $X$ is then non-compact.
In fact: 

$X^\ast$ is disconnected iff $X$ has a non-empty clopen and compact subset.

Proof: If $O$ is such a subset, $O$ is still open in $X^\ast$ by definition, $O$ will be closed in $X^\ast$ as its complement is the complement of a compact closed subset, and as such the complement is open in $X^\ast$. Hence $O$ is non-trivial (non-empty and $\infty \notin O$) clopen subset of $X^\ast$. If $X^\ast$ is disconnected so $X^\ast=U \cup V$ both open, non-empty and disjoint. Suppose WLOG that $\infty \in V$ then $X^\ast\setminus V=U$ must be closed and compact in $X$ and as $U \subseteq X$ it was also open in $X$. So $U$ is a non-empty clopen subset of $X$.
