Can $\{(f(t),g(t)) \mid t\in [a,b]\}$ cover the entire square $[0,1] \times [0,1]$ ? Suppose $f,g : [a, b] \to R$ are both continuous and of bounded variation. Then can set $\{(f(t),g(t)) \mid t\in [a,b]\}$ cover the entire square $[0,1] \times [0,1]$ ? And what if we remove the condition both $f,g$ are of bounded variation ?
 A: Claim by taking $c$ small enough we can find $x_n$ such that 
$B(x_n, c\cdot 1/n)\subset [0,1]\times [0,1]$ are pairwise distinct. Notice that the area of each ball is of order $1/n^2$, and we can imagine achieving this by separating the box into tiny boxes.
Take $t_n$ such that $f(t_n) \in B(x_n, c\cdot 1/n)$.
Now notice that for each $t_n$ you can find a  $\delta$ such that $ (t_n -\delta, t_n \delta)$ does not contain any of the other $t_i$. Therefore we can order them.
So take $\sigma: \mathbb{N} \rightarrow \mathbb{N}$ such that 
$t_{\sigma(i)}< t_{\sigma (i+1)}$.
Also now observe that $ \sum _{i=1}^{\infty}|f(t_{\sigma(i+1)}) -f(t_{\sigma(i)})|>\sum _{k=1}^{\infty} \frac{1}{k}$, since  $|f(t_i)- f(t_{\sigma(\sigma^{-1}(i)  )})|\geq 1/i$.
As mentioned before for just continuous there is a construction Peano's example.
A: We don't need to assume continuity: If $F:[0,1]\to[0,1]^2$ is a surjection, then $F$ is not of bounded variation on $[0,1].$
Proof: This will be like @clark's proof in certain ways, although I think it may be simpler. Let $n\in \mathbb N.$ Partition $[0,1]^2$ into subsquares of side-length $1/n$ in the natural way. We get $n^2$ subsquares $S_1, \dots , S_{n^2}.$  For each $k,$ let $c_k$ be the center of $S_k.$ Note that for $i\ne j, |c_i-c_j| \ge 1/n.$ Because $F$ is surjective, there exist distinct $t_k \in [0,1]$ such that $F(t_k) = c_k,$ $k=1, \dots, n^2.$ Order the $t_k$ as $t_{k_j},$ where $0\le t_{k_1} < t_{k_2} < \cdots < t_{k_{n^2}}.$ Then
$$\sum_{j= 1}^{n^2-1} |F(t_{k_{j+1}}) - F(t_{k_{j}})| =  \sum_{j= 1}^{n^2-1} |c_{k_{j+1}} - c_{k_{j}}| \ge \sum_{j= 1}^{n^2-1} \frac{1}{n} = \frac{n^2-1}{n}.$$
It follows that the total variation of $F$ on $[0,1]$ is at least $(n^2-1)/n.$ Since $n$ is arbitrary and $(n^2-1)/n \to \infty,$ we see that $F$ is not of bounded variation on $[0,1].$
