Does $(\cos t , \sin (\sqrt{2}t))$ traces all the points of a square.

It only fills the square "densely", but not completely.

Just take any $x$ coordinate you want to achieve. Then calulate one of corresponding values of $t$ for which $\cos$ achieves this value (namely $\arccos(x)$)

Then for any $k$, $t+2\pi k$ will have the same $x$ coordinate while $\sqrt{2}\times(t+2\pi k) \pmod {2\pi}$ will densely fill the $[0,2\pi]$ interval due to irrationality, so the closure is indeed the unit square.

On the other hand the point $\{1,1\}$ is never reached. Indeed for that point to be attained, the value of $t$ should be $2l\pi$ for $x$ to be correct, and $\frac{\pi+2k\pi}{\sqrt{2}}$ for the $y$ coordinate to be correct, so you need $k,l\in\mathbb{N}$ such that $$\frac{\pi+2k\pi}{\sqrt{2}}=2l\pi$$ which can be reformulated to $$\frac{\pi+2k\pi}{2l\pi}=\frac{1+2k}{2l}=\sqrt{2},$$ which would contradict the irrationality of $\sqrt{2}$.

For a fixed $x \in [-1,1]$, there are countably many $t$ that satisfy the parametric equations. Let $T$ denote such values of $t$. $y$ has period $\sqrt2\pi$.

1. Convince yourself that $T \pmod{\sqrt2\pi}$ is dense in $[0,\sqrt2\pi]$. (1, 2, 3)
2. Then, convince yourself that $\sin \sqrt 2T$ is dense in $[-1,1]$. (limit commutes with continuous function)
3. However, convince yourself that $\sin \sqrt 2T$ only has countably many numbers, while any particular vertical strip of the square has uncountably many numbers, so not all points are reached. (see comment)
4. Conclude that since $x$ is arbitrary, the set of points $(x,y)$ in the graph is dense in $[-1,1]^2$, but does not contain every point in $[-1,1]^2$.