Continuous injective functions are monotone

Idea. We want to show that the sign of $f(y)-f(x)$ is the same for all pairs $x<y$. A continuous function that never vanishes keeps a constant sign (intermediate value theorem), so we build one.

Construction. Let $x_1<y_1$ and $x_2<y_2$ be points of $I$. For $t\in[0,1]$ put
$$x_t=(1-t)x_1+tx_2,\qquad y_t=(1-t)y_1+ty_2,$$
which lie in $I$ because $I$ is an interval. We have $y_t-x_t=(1-t)(y_1-x_1)+t(y_2-x_2)>0$, so $x_t<y_t$ for all $t$.

Define $\varphi(t)=f(y_t)-f(x_t)$. It is continuous on $[0,1]$. Since $x_t\neq y_t$ and $f$ is injective, $\varphi(t)\ne0$ for all $t$.

Sign argument. A continuous function on $[0,1]$ that never vanishes has a constant sign, otherwise the intermediate value theorem would produce a zero. So $\varphi(0)=f(y_1)-f(x_1)$ and $\varphi(1)=f(y_2)-f(x_2)$ have the same sign.

Conclusion. The sign of $f(y)-f(x)$ is the same for all pairs $x<y$: either always positive (strictly increasing) or always negative (strictly decreasing).

Remark. The hypothesis "interval" is essential. On $[0,1]\cup[2,3]$ define $f(x)=x$ on $[0,1]$ and $f(x)=x-3$ on $[2,3]$ (values in $[-1,0]$). It is continuous and injective, but not monotone.