The Hausdorff dimension of the graph of $f$ equals $1$. Take a partition of $[0,1]$ by intervals of length $1/n$. Since the function is increasing you can cover the graph by boxes of size $1/n\times (f((k+1)/n)-f(k/n))$, $k=0,1,\ldots,n-1$ located above the intervals of partition. The diameters of each of the boxes is bounded by $1/n+(f((k+1)/n)-f(k/n))$ and the sum of diameters is bounded by $$ \sum_{n=0}^{n-1} \frac{1}{n}+\left(f\left(\frac{k+1}{n}\right)-f\left(\frac{k}{n}\right)\right)=2. $$ Since the diameters of the boxes covering the graph converge to $0$ as $n\to\infty$, it follows that the $H^1$ measure of the graph is bounded by $2$.
In fact the $H^1$ measure of the graph equals its length. The length of increasing function, as in your question, is always bounded from above by $2$. For the proof that, in a quite general setting, the length equals $H^1$ measure, see the book Topics on analysis in metric spaces by Ambrosio and Tilli. I will provide more detailed references when I am in my office.