I am trying to find a function $[0,1]\to[0,1]$ whose graph is dense in $[0,1]\times[0,1]$ and came up with this:
Let $f:[0,1]\longrightarrow\mathbb{R}$ be defined as follows: if $x$ is irrational then $f(x)=0$ and if $x=p/q$ is an irreducible rational then $f(x)=q$. Now $F(x)=\sin(f(x)$) is the required function (?). This are some questions that come to mind:
Let $f:\mathbb{R}\to\mathbb{R}$ be a bounded function such that $\{f(n): n\in\mathbb{N}\}\subseteq[a,b]$ is dense (in $[a,b]$) and let $g:[0,1]\to\mathbb{R}$ be another bounded function such that $g(1/n):= f(n)$. Is the segment $\{0\}\times[a,b]$ in the closure of $Graph(g)=\{(x,g(x)):x\in(0,1]\}$? I think this should be true and not so hard to prove but does the result depend on the distribution of $f(n)$ in $[a,b]$? For example, $ \sin(n)$ has density $\frac{1}{\pi\sqrt{1-x^2}}$. Does the result hold only when the distribution is uniform or whenever the density is non-zero? I'm inclined for the latter.
The idea in $\sin(f(x))$ is that $f$ blows up to infinity in every irrational and it reaches every single natural number greater than some constant $K$ (depending on the irrational) so $\{\sin(f(x)):x\text{ close to some irrational }\}=\{\sin(n):n\in\mathbb{N}\;n\ge K\}$ is still dense, which would imply that $\mathbb{I}\times[-1,1]$ (and its closure) are contained in the closure of $Graph(F)$, but its closure is $[0,1]\times[-1,1]$ because the irrationals are dense in $[0,1]$. It is like making "$\{0\}\times[-1,1]$" is in the closure of $Graph(\sin(1/x))$" but for a lot of points (a dense set of them). Does it make sense?
To find functions with dense graphs, go more extreme, e.g.'s Conway's base 13 function, or non-constructive ones that can be found by transfinite recursion and that have the property that $f[(a,b)]=\mathbb{R}$ for every $a < b$. (take the sine of it, if the image of $[0,1]$ is important)
Your proposed function $F$ is not good enough I think.