Just looking for feedback on the proof for this problem, any feedback is appreciated.
The full problem is as follows:
A normed vector space V is strictly convex if $\left\lvert\left\lvert u \right\rvert\right\rvert = \left\lvert\left\lvert v \right\rvert\right\rvert = \left\lvert\left\lvert \frac{u + v}{2} \right\rvert\right\rvert = 1$ for vectors $u,v \in V$ implies that $u=v$
Show that an inner product space is always strictly convex.
Proof:
Let $\left\lvert\left\lvert u \right\rvert\right\rvert = \left\lvert\left\lvert v \right\rvert\right\rvert = \left\lvert\left\lvert \frac{u + v}{2} \right\rvert\right\rvert = 1$
Therefor:
$\left\lvert\left\lvert u \right\rvert\right\rvert = \left\langle u,u \right\rangle^{1/2} = 1 \Leftrightarrow \left\langle u,u \right\rangle = 1$
$\left\lvert\left\lvert v \right\rvert\right\rvert = \left\langle v,v \right\rangle^{1/2} = 1 \Leftrightarrow \left\langle v,v \right\rangle = 1$
$\left\lvert\left\lvert \frac {u+v}{2} \right\rvert\right\rvert = \left\langle \frac {u+v}{2},\frac {u+v}{2} \right\rangle^{1/2} = 1 \Leftrightarrow \left\langle \frac {u+v}{2},\frac {u+v}{2} \right\rangle = 1$
Now,
$\langle \frac{u + v}{2}, \frac{u + v}{2} \rangle = \frac{1}{4} \langle u, u \rangle + \frac{1}{2} \langle u, v \rangle + \frac{1}{4} \langle v, v \rangle = 1$
$\to \frac{1}{2} + \frac{1}{2} \langle u, v \rangle = 1$
$\to \langle u,v \rangle = 1$
Next, $\langle u - v, u - v \rangle = \langle u,u \rangle - 2\langle u,v \rangle + \langle v,v \rangle = 1 - 2(1) + 1 = 0$
This implies that $u - v = 0$, or that $u = v$.
Q.E.D.
Your proof is correct if $V$ is an inner product space over the real numbers $\Bbb R$. If the underlying field is $\Bbb C$ then $$ \begin{align} \langle u - v, u - v \rangle &= \langle u,u \rangle - \langle u,v \rangle- \langle v,u \rangle + \langle v,v \rangle \\ &= \langle u,u \rangle - \langle u,v \rangle- \overline{\langle u,v \rangle} + \langle v,v \rangle \\ &= \langle u,u \rangle - 2 \operatorname{Re}\bigl(\langle u,v \rangle \bigr) + \langle v,v \rangle \end{align} $$ and similarly for the expansion of $\langle \frac{u + v}{2}, \frac{u + v}{2} \rangle$. With that adjustment, your proof can be made to work for the general case.
But actually the value of $\langle u,v \rangle+ \langle v,u \rangle$ does really matter because it cancels when we add the equations $$ \langle u + v, u + v \rangle = \langle u,u \rangle + \langle u,v \rangle+ \langle v,u \rangle + \langle v,v \rangle \, ,\\ \langle u - v, u - v \rangle = \langle u,u \rangle - \langle u,v \rangle- \langle v,u \rangle + \langle v,v \rangle \, . $$ This gives the parallelogram law $$ \Vert u+v\Vert^2 + \Vert u-v\Vert^2 = 2 \Vert u\Vert^2+2 \Vert v\Vert^2 $$ which holds in every (real or complex) inner product space, and the strict convexity follows immediately: If $\Vert u\Vert = \Vert v\Vert = \Vert \frac{u+v}{2}\Vert = 1$ then $$ \Vert u-v\Vert^2 = 2 \Vert u\Vert^2+2 \Vert v\Vert^2-\Vert u+v\Vert^2 = 2 + 2 - 4 = 0 $$ and therefore $u=v$.