Is the following argument for the given proof correct?
Suppose $u,v\in V$. Prove that $\|au+bv\| = \|bu+av\|$ for all $a,b\in\mathbb{R}$ if and only if $\|u\| = \|v\|$.
Proof. $(\Rightarrow)$. Assume that $\|au+bv\| = \|bu+av\|$ for all $a,b\in\mathbb{R}$ then in particular for $a = 1$ and $b=0$ we have $\|u\| = \|v\|$.
$(\Leftarrow)$. Let $a$ and $b$ be arbitrary real numbers and assume that $\|u\| = \|v\|$ consequently $|a|^2\cdot\|u\|^2 = |a|^2\cdot\|v\|^2$ and $|b|^2\cdot\|v\|^2 = |b|^2\cdot\|u\|^2$ adding these two equations yields $|a|^2\cdot\|u\|^2+|b|^2\cdot\|v\|^2 = |a|^2\cdot\|v\|^2+|b|^2\cdot\|u\|^2$ equivalently $\|au\|^2+\|bv\|^2 = \|av\|^2+\|bu\|^2$ equivalently $\langle au,au\rangle+\langle bv,bv\rangle = \langle av,av\rangle+\langle bu,bu\rangle$ and by applying additivity in first slot and then additivity in second slot we have $\langle au+bv,au+bv\rangle = \langle bu+av,bu+av\rangle$ equivalently $\|au+bv\|^2 = \|bu+av\|^2$ taking square root yields the required result.
$\blacksquare$
Yes, both directions are correct.