Let 𝐴 be a connected subspace of ℝ that is not an interval. Then there exist 𝑎,𝑏 ∈𝐴 such that 𝑎 <𝑥 <𝑏 for some 𝑥 ∉𝐴. Then 𝐴 may be partitioned into two disjoint open sets as follows

contradicting our requirement that 𝐴 be connected.

Conversely, let 𝐼 be an interval and 𝐼 =𝑈 ∪𝑉 for some inhabited disjoint open 𝑈,𝑉. Without loss of generality assume there exists 𝑥 ∈𝑈,𝑦 ∈𝑉 such that 𝑥 <𝑦. By the completeness of ℝ, the supremum 𝑠 =sup𝑈 ∩[0,𝑦) exists, and 𝑥 <𝑠 ≤𝑦, so either 𝑠 ∈𝑈 or 𝑠 ∈𝑉, and from openness (𝑠 −𝛿,𝑠 +𝛿) is either a subset of 𝑈 or 𝑉. Both situations are a contradiction.