If I want show that the Hilbert-Curve is well-defined, I think that it is sufficient to show, that $(H_n(x))_{n \in \mathbb{N}}$ is a Cauchy-sequence for every $x \in [0,1]$, where $H_n$ is the n-th Hilbert-Polygon.
If I am not wrong, than this property follows directly from the fact, that for any $x \in [0,1]$ the sequence doesn't leave its "square". Take $H_n(x)$, than $H_{n+i}(x), i \in \mathbb{N}$ will be in the same square with sidelength $\frac{1}{2^n}$ where $H_n(x)$ is located.
From here, there are no difficulties to write that down formally. I only wanted to check if my idea/argument is right.