I found this marvellous thing while playing around in Desmos!
It seems to me that the curve $\vec{r}(t) = \left[ \cos ( t^2 ), \sin t \right] \left( t \in [0, +\infty) \right)$ is filling a square. But, as @BarryCipra kindly pointed out, it crosses the X axis only when $t$ is an integer multiple of $\pi$, so it crosses the X axis countably many times. And we know that $[-1, 1]$ is uncountable, so it seems the curve doesn't actually fill the square. A question that would be fair to ask: Does the curve get arbitrarly close to every point in the square? Or: Is $[-1, 1]^2$ the infinite closure of the curve?
Intuitively, as $t$ gets larger and larger, we see that $\cos(t^2)$ oscillates faster and faster relative to $\sin(t)$, so we can choose very large $t$ that gives the right $y$-value and then just perturb it a bit to get the right $x$-value while only changing $y$ by a very small amount.
Let's make this more precise. Let $(x_0, y_0) \in [-1, 1]^2$ and $\varepsilon > 0$ be arbitrary. Let $t_0 \geq \frac{\pi}{\varepsilon}$ such that $\sin(t_0) = y_0$. We have $\cos(t_0^2 + c) = x_0$ for some $c \in [0, 2\pi]$. Let $t = \sqrt{t_0^2 + c}$. Then $\cos(t^2) = x_0$. Moreover, $$\lvert \sin(t) - y_0 \rvert = \lvert \sin(t) - \sin(t_0) \rvert \leq \lvert t - t_0 \rvert = \sqrt{t_0^2 + c} - t_0$$ $$ \mathrel{\leq} \sqrt{t_0^2 + 2\pi} - t_0 = \frac{2\pi}{\sqrt{t_0^2 + 2\pi} + t_0} < \frac{\pi}{t_0} \leq \varepsilon.$$ Thus, the distance between $(\cos(t^2), \sin(t))$ and $(x_0, y_0)$ is less than $\varepsilon$.
In particular, the curve gets within distance $\varepsilon$ of every point in $[-1, 1]^2$ as $t$ ranges over the interval $[0, \frac{\pi}{\varepsilon} + 2\pi]$. (I'm not claiming this bound is necessarily optimal, to be clear, but I'd guess it's at least pretty close.)