 Solution

The intermediate value theorem states that if $f:[a,b]\to \mathbb{R}$ is continuous and $y$ lies between $f(a)$ and $f(b)$, then $f(c)=y$ for some $c\in [a,b]$.

Let $x_1<x_2<x_3$. If, for example, $f(x_2)$ exceeded both $f(x_1)$ and $f(x_3)$, choose $y$ strictly between $f(x_2)$ and $\max \{f(x_1),f(x_3)\}$. Applying the theorem on both $[x_1,x_2]$ and $[x_2,x_3]$ would give two distinct preimages of $y$, contradicting injectivity. The analogous argument excludes $f(x_2)$ below both endpoint values. Since the three values are distinct,

Fix $a<b$. If $f(a)<f(b)$, applying the displayed betweenness property to triples containing $a,b$ forces the same increasing order for every pair $x<y$; an order reversal would create a triple whose middle value is not between the other two. Thus $f$ is strictly increasing. If $f(a)>f(b)$, the same argument shows that it is strictly decreasing. This proves that every continuous bijection of the real line is monotone.