Can $X$ have compact connected components? Let $X$ be a Hausdorff, locally compact but non-compact topological space.
If the (Alexandroff) one-point compactification is connected, can $X$ have compact connected components?
I think I proved the following


  Lemma
  
  Let $X$ be a Hausdorff space and $C \subset X$ have a compact neighbourhood $K$. Then $C$ is a component of $X$ if and only if $C$ is a component of $K$.


in this answer.

For the present problem this implies a negative answer, since a compact set
in a locally compact Hausdorff space has a compact neighbourhood. Let $C$ be
a compact component of $X$, and let $K$ be some compact neighbourhood of $C$ in
$X$. Applying the lemma one way then shows that $C$ is a component of $K$, and
then applying it the other way shows that $C$ is a component of the
compactification, since $K$ remains a compact neighbourhood of $C$.
Well, no. Assume $X=Y\sqcup C$, where $C$ is compact. Then the 1PC is $X\cup\{\infty\}$ with the topology: $U$ is open iff $U\subset X$ is open in the previous topology or $\infty\in U$ and $U^c$ is compact. Then $C = (U\cup\{\infty\})^c$ is compact, and thus $U\cup\{\infty\}$ is open. $C$ is also open (it was open in $X$). Thus the 1PC of $X$ is not connected.