Closed Interval in Reals is Uncountable

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

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

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

We have that the unit interval is uncountable.

Let $f : [0,1] \to [a,b]$ such that $fx = a + (b - a)x$.

Then if $fx_1 = fx_2$, 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 $[a,b]$ were countable, there would be an injective function $g : [a,b] \to \N$.

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

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

We know by the above that $[c,d]$ is uncountable.

Let $S = \{x \in \R : a \leq x \leq b\}$.

The identity function $\operatorname{id} : [c,d] \to S$ 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 $[c,d] \to \N$ given by $g \circ \operatorname{id}$.

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