Is the one-point compactification of a connected and locally connected space also locally connected?
My guess is no, because I haven't been able to prove it. But of course I also haven't come up with a counterexample...
Here is my proof attempt:
Let $X$ be a connected, locally connected, locally compact space. Let $X'=X\cup\{\infty\}$ be the one-point compactification of $X$. We show $X'$ is locally connected at $\infty$. Suppose not. Let $U$ be an open set in $X'$ such that $\infty\in U$ and no connected open subset of $X'$ containing $\infty$ is contained in $U$. We can partition $U$ into disjoint open sets, then partition the one containing $\infty$ into disjoint open sets, etc. By compactness, the boundary of $U$ should be covered by finitely many of the open sets which miss $\infty$. But since $X'$ is compact and connected, the component of $\infty$ in $U$ meets the boundary of $U$; this component goes through disjoint open sets and so we have a contradiction.
Note: This answer was for the original version of the question, which did not require the space to be connected.
A discrete space is locally connected; its one point compactification is not, assuming that the discrete space is infinite.