Compact Connected Hausdorff Space has no compact component in the complement of a point Let $X$ be a Hausdorff Compact Connected Space. Prove that $X\setminus\{x\}$ can't be expressed by the disjoint union of two connected sets with one them being compact.(lets assume the empty set is connected)
So, I tried a demonstration by contradiction:
If $X\setminus\{x\}$ is connected, then if $X\setminus\{x\}$ is compact, it's a closed subspace of $X$ (because $X$ is Hausdorff) and therefore it's a clopen on $X$. So, $X\setminus\{x\}$ can't be disconnected. Assume that there are disjoint connected sets $U$ and $V$ of $X\setminus\{x\}$ such that  $$X\setminus\{x\} = U \cup V$$
And $U$ is a compact set. $U$ can't be open on $X\setminus \{x\}$ (or otherwise it'd be a clopen set of $X$). But as $U \cup V =X= \cup_{i\in I} A_i$ with $A_i$ open connected subsets of $X\setminus \{x\}$. Because $U$ is connected, $U \subset A_{i_1}$ and the same goes to $V \subset A_{i_2}$. $X \subset U \cup V \subset A_{i_1} \cup A_{i_2}$. Because $X$ is not connected, $i_1 \neq i_2$. Also, because $A_{i_2} \cap U = \emptyset$ and then $A_{i_2} = V$ and therefore $U = A_{i_1}$ and $U$ is open, a contradiction.

As it is stated, the proposition is false (check Brian's answer). I've corrected it in another question: Complement of a point of a Compact Connected Hausdorff Space has no compact maximal connected subspace


Revised and corrected to match the edited question.
The statement is false. Let $X=[0,2]$ with the usual topology, and let $x=0$. Then 
$$X\setminus\{x\}=(0,2]=(0,1)\cup[1,2]\;,$$
where $(0,1)$ and $[1,2]$ are disjoint and connected, and $[1,2]$ is compact.
