real analysis - Injective functions with intermediate-value property are continuous. Better proof?

A function $f: \mathbb{R} \to \mathbb{R}$ is said to have the intermediate-value property if for any $a$, $b$ and $\lambda \in [f(a),f(b)]$ there is $x \in [a,b]$ such that $f(x)=\lambda$.

A function $f$ is injective if $f(x)=f(y) \Rightarrow x=y$.

Now it is the case that every injective function with the intermediate-value property is continuous. I can prove this using the following steps:

1. An injective function with the intermediate-value property must be monotonic.


2. A monotonic function possesses left- and right-handed limits at each point.


3. For a function with the intermediate-value property the left- and right-handed limits at $x$, if they exist, equal $f(x)$.

I am not really happy with this proof. Particularly I don't like having to invoke the intermediate-value property twice.

Can there be a shorter or more elegant proof?