Direct proof for intervals in $\mathbb{R}$ Let $S$ be a non empty connected subset of $\mathbb{R}$. Prove that $S$ is an interval.
I have seen many proofs of this using contradiction. I'm just wondering if there is a direct proof for it without using contradiction.
 A: In case $S \subset \mathbb{R}$ is open (without its endpoints, which can be added later once we get the open interval). We can show, directly, that $S$ being connected in $\mathbb{R}$ $\implies$ path-connected.
Pick any $x_0$ in $S$ and define $P = \{x \in S: \text{ there is a continuous path from } x_0 \to x \}$. We show that $P$ and $S \setminus P$ are both open, but as $S$ is connected, $S = P$, i.e. is path connected. 
Let $x_1 \in P$. Then $\exists \epsilon: (x_1 - \epsilon, x_1 + \epsilon) \subset S$ since $S$ is open, and if $x_2 \in (x_1 - \epsilon, x_1 + \epsilon)$ we can find a path from $x_0 \to x_1 \to x_2$. Hence $P$ is open.
Let $x'_2 \in S \setminus P$. Then $\exists \epsilon: (x'_2 - \epsilon, x'_2 + \epsilon) \subset S$. If $x'_1 \in  (x'_2 - \epsilon, x'_2 + \epsilon) \cap P$, then there would be path $x_0 \to x'_1 \to x'_2$, but $x'_2 \notin P$. Therefore, $(x'_2 - \epsilon, x'_2 + \epsilon) \in S \setminus P$. QED.
Now, taking the open interval between infimum and supremum  of $S$ (or $\pm \infty$) and showing inclusion, in both directions, will finish the proof.
