Let $f:[0,1]\to\mathbb{R}$ be continuous. We define the graph of $f$ to be the set $$\text{graph}\,f=\{(x,f(x)):x\in [0,1]\}.$$
Is $\mathcal{H}^1(\text{graph}\,f)\ge 1$, where $\mathcal{H}^1$ is the one-dimensional Hausdorff measure on $\mathbb{R}^2$?
I know that if $f$ is Lipschitz continuous, then it is differentiable almost everywhere and $$\mathcal{H}^1(\text{graph}\,f)=\int_0^1\sqrt{1+(f'(x))^2}dx\ge 1.$$
I believe that the result still holds in the case where $f$ is just continuous but I don't know how to show this.
In fact, $f$ doesn't even need to be continuous. Hint: Consider the projection $\pi_1: (x,y) \to x$ which maps the graph onto $[0,1]$.