What is the proof that SVM can be used to solve the least squares problem with norm equality constraint?

Norm $\| \cdot \|$ is invariant under unitary transformation so:

$$\|Ab\| =\| U\Sigma V^* b\| = \|\Sigma b'\|$$

Where $b' = V^* b$, so $\|b'\| = \|V^* b\| = \|b\| = 1$.

Next we have that:

$$\text{argmin}_b \|\Sigma V^* b\| = V\text{argmin}_{b'} \| \Sigma b' \|$$

This is because $V^*$ maps unit sphere onto unit sphere.

And that $b'$ which minimizes $\|\Sigma b'\|$ is $(0,\dots,0,1)^T$.

Finally $V (0,\dots,0,1)^T$ is equal to the last column of $V$.

Note that if $A = U\Sigma V$, then $A^* A = V^*\Sigma^*U^*U\Sigma V = V^*\Sigma^*\Sigma V$. Therefore, the eigenvalues of $A^*A$ are equal to square of the absolute values of the singular values of $A$. Also the minmization problem can be seen as $$ \underset{b: \;||b||=1}{argmin} \; ||Ab||= \underset{b: \;||b||=1}{argmin}\;||Ab||^2 = \underset{b: \;||b||=1}{argmin}\; b^*(A^*A) b $$ This is equivalent to finding the eigenvector corresponding to the minimum eigenvalue of $A^TA$, which is precisely the column in $V$ corresponding to the minimum eigenvalue in the diagonal matrix $\Sigma^*\Sigma$ (which is in turn the absolute square of the minimum singular value of $A$).