Let $V$ be a inner product space. Given that $S=\{u\in V:||u||=1\}$ and let $0\ne v\in V$, prove that $$\left\|v-\frac{v}{\|v\|}\right\|= \min\{\|v-u\|:u\in S\}$$
Here is my proof so far:
According to inner product identities:
$$\|v-u\|^2=\|v\|^2-2Re(\langle v,u\rangle)+\|u\|^2.$$
We can use Cauchy-Schwarz to show that:
$$Re(\langle v,u\rangle)\le \|v\|\cdot\|u\|.$$
So we can determine that:
$$\|v\|^2-2\|v\|\cdot\|u\|+\|u\|^2\le\|v\|^2-2Re(\langle v,u\rangle)+\|u\|^2.$$
Because $\|u\|=1$:
$$\|v\|^2-2\|v\|+1\le\|v\|^2-2Re(\langle v,u\rangle)+1.$$
According to the same identities:
$$\left\|v-\frac{v}{\|v\|}\right\|^2=\|v\|^2-2Re\left(\left\langle v,\frac{v}{\|v\|}\right\rangle\right)+\left(\left\|\frac{v}{\|v\|}\right\|\right)^2.$$
Since $$\left\langle v,\frac{v}{\|v\|}\right\rangle=\frac{1}{\|v\|}\langle v,v\rangle=\sqrt{\langle v,v\rangle}=\|v\|,$$ therefore $$\left\|v-\frac{v}{\|v\|}\right\|^2=\|v\|^2-2\|v\|+1\le\|v\|^2-2Re(\langle v,u\rangle)+1=\|v-u\|^2.$$
Hence, $$\left\|v-\frac{v}{\|v\|}\right\|^2\le\|v-u\|^2 \text{ and }\left\|v-\frac{v}{\|v\|}\right\|\le\|v-u\|.$$
Your proof is correct except for a minor typo after applying Cauchy-Schwarz:
$$||v||^2-2||v||\cdot||u||+||u||^{\large\color{red}{2}}\le||v||^2-2Re(\langle v,u \rangle)+||u||.$$