Compact connected space is the union of two disjoint connected sets. This was mentioned in an article, but I have not been able to find a proof anywhere.  Assume the original space has at least 2 points. The disjoint connected sets should be nontrivial (each has at least 2 points). 
For instance $[0,2]=[0,1]\cup (1,2]$.
I tried to use the boundary bumping theorem.
 A: Here's my attempt.  Please let me know if it looks ok.
Let $X$ be a compact connected space with at least two points. By the boundary bumping theorem there is a proper closed connected $K\subseteq X$ with at least 2 points.  We may assume $X\setminus K$ is not connected, say $X\setminus K=U\cup V$. Let $p\in U$ and let $C$ be the component of $p$ in $U$.  By the boundary bumping theorem $C$ is nontrivial. For each $q\in U\setminus C$ let $C_q$ be the component of $q$ in $U$. By the boundary bumping theorem each $C_q$ limits to $K$. Now we have $C$ and $K\cup V\cup \bigcup _{q\in U\setminus C} C_q$ are disjoint nontrivial connected sets unioning to $X$.
Note: I used the classic lemma: If $X$ is connected, $K\subseteq X$ is connected, and $X\setminus K=U\cup V$, then $K\cup U$ and $K\cup V$ are connected.
Simplification suggested by @Daron: Let $C$ be one component of $X\setminus K$; $C$ is nontrivial. Then $X\setminus C$ is also connected and nontrivial and as it is equal to $K$ with the potential addition of some components of $X\setminus K$ (these components limit to $K$).
A: I cannot imagine this theorem holds for all topological spaces. Nevertheless, if we restrict to spaces where the boundary-bumping theorem holds (usually compact, connected, Hausdorff spaces) then we can do the following.
Choose any point $x \in X$. Supposing if $(X-x)$ is connected we can write $X = \{x\} \cup (X-x)$ as required. Otherwise $(X-x)$ is partitioned into connected components. For each component $F$ of $(X-x)$ the closure $\overline F = F \cup \{x\}$ by boundary-bumping. Moreover the closure is a connected set. So split the set of connected components of $(X-x)$ into two nonempty parts $\mathcal F$ and $\mathcal G$ and consider the two sets
$A =\bigcup  \{\overline F \colon F \in \mathcal F\}$
$B =\bigcup  \{\overline G \colon G \in \mathcal G\}$ 
Every $\overline F$ includes the point $x$ and likewise for evry $\overline G$. So $A$ and $B$ are connected. By construction we have $X = A \cup B$ as required. 
Edit: If you want the intersection of $A$ and $B$ to be nontrivial, then you can replace $\{x\}$ with any proper subcontinuum $K$ and repeat the proof. That proper subcontinua always exist is a well-known corollary to the boundary-bumping theorem. All you need to change is that, instead of  $\overline F = F \cup \{x\}$ we have that $\overline F$ meets $K$ and hence $\overline F \cup K$ is connected (and closed). Then use the sets
$A =\bigcup  \{\overline F \cup K \colon F \in \mathcal F\}$
$B =\bigcup  \{\overline G \cup K \colon G \in \mathcal G\}$ 
