Is there a bijective function $f:[0,1] \to [0,1]$ such that the graph of $f$ in $\mathbb{R}^2$ is a dense subset of $[0,1] \times [0,1]$?

Yes, what you can do is construct an injective function $f:\mathbb Q \cap [0,1] \rightarrow \mathbb [0,1]$ whose graph is dense in $[0,1] \times [0,1]$ and then extend the domain of $f$ to $\mathbb [0,1]$ in a way that makes $f$ a bijection (this is doable since there are $|\mathbb R | $ points in $[0,1]$ not already in the image of $f$).

For example, On $\mathbb Q \cap [0,1]$ you could let $$f \left ( \frac{a}{b} \right ) = \frac{\pi a^2}{b} \mod 1$$