Prove that a strongly convex function imples $2c(F(w)-F_*) \leq ||\nabla F(w)||_2^2$ 
The proof is given as follows:

My question is why is the unique minimizer $\bar{w}_* = w - \frac{1}{c} \nabla F(w)$?

There is an easier argument for quadratic functionals: completing the square. The following identity is just as easy to prove as the 1D version you know from highschool:
$$ \alpha ‖x‖^2 + \beta · x = \alpha\left\|x+\frac1{2\alpha}{\beta} \right\|^2 -  \frac{‖\beta‖^2}{4\alpha}$$
In your case, this yields (remember, $w$ is fixed, $\bar w$ is varying) with $$x = w - \bar w, \quad \alpha = \frac{c}{2},\quad  \beta = \nabla F(w)$$
$$⟹ q(\bar w) = C(w) + \frac c{2}\left\|\bar w - \left(w-\frac1c∇ F(w)\right)\right\|^2 $$
where $C(w) = F(w)-\frac{‖∇ F(w)‖^2}{2c}$. Its now obvious what the minimiser and minimum is.
