 Solution

If $x\ne y$, one of $x<y$ and $y<x$ holds. Since $f$ is strictly increasing, the corresponding values are strictly ordered, so $f(x)\ne f(y)$. Thus $f$ is an injection. Every element of the image of a function $f([a,b])$ equals $f(x)$ for some $x\in [a,b]$, which is precisely surjectivity onto that image of a function. Hence $f:[a,b]\to f([a,b])$ is a bijection.