One way to interpret $\nabla f$ geometrically is that it yields a vector field who's direction is in the direction of the steepest ascent.
So, we can find the unit vector along this vector:
$$\frac{ \nabla f}{ || \nabla f || }$$
If we multiply it by the scaler $-1$:
$$ -\frac{ \nabla f }{ || \nabla f || } $$
Would this give me the unit vector along the steepest descent?
I needed the unit vector along steepest descent. After thinking a bit I came up with this but I am not sure if this is correct. In case it is correct, is there any proofy way to explain it so I can be more confident about it?
Your answer is correct.
To measure the extent of ascent/descent along a unit vector, you need the directional derivative. For any differentiable $f$, the directional derivative along a unit vector $u$ is given by $$ D_uf = \nabla f \cdot u $$ Note, however, that by the Cauchy Schwarz inequality (or the "cosine formula" for the dot product), $$ |D_uf| = |\nabla f \cdot u| \leq \|\nabla f\| \cdot \|u\| = \|\nabla f\| $$ So, the lowest $D_uf$ can be is $- \|\nabla f\|$. It suffices then to confirm that setting $u = -\nabla f/\|\nabla f\|$ yields this lowest possible directional derivative, namely $D_u f = -\|\nabla f\|$ (as desired).