Interval Divided into Subsets

Theorem
Let $$\mathbb{I}$$ be a real interval.

Let $$S$$ and $$T$$ be non-empty subsets of $$\mathbb{I}$$ such that $$\mathbb{I} = S \cup T$$.

Then one of $$S$$ or $$T$$ contains an element at zero distance from the other.

Proof
$$\mathbb{I} = S \cup T \Longrightarrow \forall x \in \mathbb{I}: x \in S \lor x \in T$$ from the definition of union.

That is, every element of $$\mathbb{I}$$ belongs either to $$S$$ or to $$T$$.

The distance of a point $$c \in \mathbb{R}$$ from a subset $$S$$ of $$\mathbb{R}$$ is given as $$d \left({c, S}\right) = \inf_{x \in S} \left({\left|{c - x}\right|}\right)$$.

Assume that $$S$$ and $$T$$ have no elements in common, otherwise the result is trivial.

Suppose that $$\exists s \in S, t \in T$$ such that $$s < t$$.

(If not, then $$\exists s \in S, t \in T$$ such that $$s > t$$, and the following argument may be amended appropriately.)

Let $$T_0 = \left\{{x: x \in T: x > s}\right\}$$.

As $$t \in T_0$$ it follows that $$T_0 \ne \varnothing$$. Also, $$T_0$$ is bounded below by $$s$$.

Let $$b = \inf \left({T_0}\right)$$.

If $$b \notin T$$ then $$b \in S$$.

But from Distance from Subset of Real Numbers‎, it follows that $$d \left({b, T_0}\right) = 0$$.

Thus we have found a point in $$S$$ which is zero distance from $$T$$.

Otherwise, $$b \in T$$.

Then $$b > s$$, and the open interval $$\left({s \, . \, . \, b}\right)$$ is a non-empty subset of $$S$$.

Hence $$b$$ is a point in $$T$$ which is zero distance from $$S$$.