Separation of two points with null-sets

I am asked the following question: (I write $\{a,b\}$ for points in $\mathbb{R}^2$)

Let $\{0,0\},\{1,1\} \notin K \subset [0,1]^2$ such that the projections of $K$ onto the $x$-axis and the $y$-axis are (1 dimensional) lebesgue null-sets. Is there a curve $\gamma : [0,1] \longrightarrow [0,1]^2\backslash K$ such that $\gamma(0)=\{0,0\}, \gamma(1)=\{1,1\}$ and $\ell(\gamma)\leq2$?

My idea was to consider a family of disjunct curves to do a dimension argument, but I stumbled across the cantor-set and therefore saw that $K$ doesn't have to be countable, might there even be a counter example?

I believe you can construct such a curve $\gamma$ for all such sets $K$ by zig-zagging in axis-parallel lines. Let $\pi_x$ and $\pi_y$ be the projections of $K$ onto the $x$ and $y$ axis, respectively. Pick some starting point $(x_0,y_0)$ with $y_0\notin\pi_y$, and pick some $x_1\notin\pi_x$ with $x_1\le x_0/2$. Add the segment from $(x_1,y_0)$ to $(x_0,y_0)$ to the curve (say, by mapping $[\frac14,\frac12]$ to it). Then pick some $y_1\notin\pi_y$ with $y_1\le y_0/2$ and add the segment from $(x_1,y_1)$ to $(x_1,y_0)$ to the curve (say, by mapping $[\frac18,\frac14]$ to it). Repeating this procedure indefinitely constructs a continuous curve $\gamma$ that can be continuously extended to $\gamma(0)=(0,0)$, since the $x_i$ and $y_i$ converge to $0$. We can do the same thing in the other direction towards $(1,1)$. The length of the curve is $2$ as required.