Closed Interval in Reals is Uncountable

Theorem
Let $a, b$ be extended real numbers such that $a < b$.

Then the set $\left\{{x \in \R : a \le x \le b}\right\} \subseteq \R$ is uncountable.

Proof
First suppose that $a, b \in \R$.

We have that the unit interval is uncountable.

Let $f : \left[{0 .. 1}\right] \to \left[{a .. b}\right]$ such that $f \left({x}\right) = a + (b - a)x$.

Then if $f \left({x_1}\right) = f \left({x_2}\right)$, we have $a + (b-a) x_1 = a + (b-a) x_2$.

Since $b > a$, $b - a > 0$, so this implies $x_1 = x_2$.

Therefore $f$ is injective.

Now if $\left[{a .. b}\right]$ were countable, there would be an injective function $g : \left[{a .. b}\right] \to \N$.

By Composite of Injections is an Injection, this implies that $g \circ f$ is an injection, so $\left[{0 .. 1}\right]$ is countable, a contradiction.

If one or both of $a, b$ is (are) not real, then we can pick a closed interval $\left[{c .. d}\right] \subseteq \left[{a .. b}\right]$ with $c, d \in \R, c < d$.

We know by the above that $\left[{c .. d}\right]$ is uncountable.

Let $S = \left\{{x \in \R : a \le x \le b}\right\}$.

The identity function $\operatorname{id} : \left[{c .. d}\right] \to S$ is trivially injective.

If $S$ were uncountable, we would have an injective function $g : S \to \N$.

Then by Composite of Injections is an Injection we would have an injection $\left[{c .. d}\right] \to \N$ given by $g \circ \operatorname{id}$.

This was shown above to be false, so $S$ is uncountable.