Are the closed intervals of $\mathbb{R}$ precisely the compact connected sets? Equip $\mathbb{R}$ with the topology generated by open intervals $(a, b)$. A subset of $\mathbb{R}$ is compact iff it's closed and bounded.
Is every closed bounded connected subset of $\mathbb{R}$ a closed interval $[a, b]$ (and conversely)?
Is every open bounded connected subset of $\mathbb{R}$ an open interval $(a, b)$ (and conversely)?
Is this somehow related to the fact that removing one point from $\mathbb{R}$ splits it into 2 disconnected pieces (how is this property called anyway)?
 A: Clearly a closed and bounded interval is compact and connected.
Conversely, if a set is connected, then it is an interval, meaning it is a set $I$ with the following property: for all $x,y \in I$, if $x < z<y$, then $z \in I$. All the sets with this property must have one of the forms $$\{a\},\,[a,b],\, ]a,b[, \,[a,b[, \,]a,b], \,]a,+\infty[,\, [a,+\infty[,\, ]-\infty,b[ \mbox{ or } ]-\infty,b].$$
Among these, only $\{a\}$ and $[a,b]$ are compact, hence the answer to your question is yes. Notice that $\{a\}$ is an interval, by definition.
A: Lemma 1:  Only intervals and singletons and the empty set are connected and all intervals and singletons and the empty set are are connected.
Lemma Z: $K\subset \mathbb R$ is a interval if and only if for all $x,y \in K$ then for all $k; x < k < y; k \in K$.
Proof:  Should be self-evident.  If $K$ is an interval than $K = [(a,b)]$ (for sake of notation $a$ can be $-\infty$ and $b$ can be $\infty$). and $a \le x < y \le b$ and for all $k: x < k < y$ then $a < k < y$ so $k \in K$. 
If there exists a $k$ so that $x < k < y$ with $k \not \in K$ and $x,y \in K$ then there is no $a,b$ (not even $\pm \infty$) so that $a \le x; b \ge y$ and for all $r \in \mathbb R$ $a < r < b$; $r \in K$. (as $a < k < b$ but $k \not \in K$).  So $K$ would not be an interval.
Proof of Lemma 1: If $K$ = $\emptyset$ or $K = \{x\}$, some singleton then $K$ can not be partitioned into two partitions so $K$ is connected.
Let $E \subset \mathbb R$.  And let $E$ be such that there exist $x,y\in E; x< y $ and there exists a $k \not \in \mathbb R$ so that $x < k < y$.  Let $A = E \cap (\infty, k)$ and $B= E \cap (k, \infty)$ then $A, B$ are non-empty partitions of $E$ and $\overline A \subset (-\infty,k]$ is disjoint from $B \subset (k,\infty)$ and $\overline B \subset [k,\infty)$ is disjoint from $A \subset (-\infty, k)$ so $E$ is not connected.
So the only connected subsets of $E$ are intervals, singletons, and the empty set.
If $K$ is an interval and $K = A\cup B$ and $A,B$ non empty and $A \cap B = \emptyset$.  Let $a \in A$ and $b \in B$ and wolog $a < b$.  Let $K = \{x| x \ge a; \forall k; a\le k \le x: k \in A\}$ and let $L = \{y \in K| y > a; y \in B\}$.  It's easy to prove $K$ is non-empty $(a \in K)$ and bounded above (by $b$) and $ L$ is non-empty ($b$ is in it) and bounded below (by $a$) and that $\sup L = \inf L$. As $K$ is an interval and $a \le j = \sup K = \inf L\le b$ then $j \in K$.  And $j \in \overline A$ and $k \in \overline B$.  So either $j \in \overline A \cap B$ or $j \in \overline B \cap A$.  So $K$ is connected.
