To introduce some context, here I am dealing with a Hilbert space H and a linear and closed subspace $S$ of $H$ such that $(b_n)_{n \in \mathbb N}$ is an orthonormal basis for $S$. Previously, I have shown that the vector $$ y = \sum_{k=1}^\infty \langle x,b_k\rangle b_k$$ Satisfies the equality $\| x- y\| = \inf_{z \in S} \| x- z \|.$ I am now trying to prove that this $y$ is unique.
To do so, my thinking go as follows:
Suppose there exists a distinct element $y' \neq y$ ($y' \in S$) such that $\| x - y'\| = \inf_{z \in S} \| x - z \| = \|x-y\|.$ To prove what's wanted, one should find a contradiction (essentialy, should find that $y = y'$). My only idea to reach such a contradiction is to apply the parallelogram law to the vectors $x-y$ and $x-y'.$ Doing so,
$$ \| x - y\ + x - y'\|^2 + \| (x-y)-(x-y') \|^2 = 2(\|x-y\|^2 + \|x-y'\|^2) $$
But it is easy to see that $x-y+x-y' = 2x - y - y'$ and $(x-y)-(x-y') = y' - y$ and $\|x-y'\|^2 = \|x-y\|^2$, by hipothesis. Thus, the equation above is equivalent to
$$ \| 2x-y-y'\|^2 + \|y' - y\|^2 = 4\|x-y\|^2 \Leftrightarrow \| y' - y \|^2 = 4\|x-y\|^2 - \|2x-y-y'\|^2. $$
Now, the desired result would come if I could prove that $4\|x-y\|^2 = \|2x-y-y'\|^2$ but I don't see how to do this...
Any hint is appreciated.
A few comments first:
If you prove that $y=y'$ you already got what you want; phrasing it as a contradiction only obscures the argument.
Nowhere in your argument do you use that $y,y'$ minimize the distance.
And, also, nowhere in your argument do you use that $S$ is convex; and the uniqueness can fail for non-convex $S$.
With the above considered, a typical argument is as follows. It uses the paralellogram identity in a similar way as you did. The crucial piece of information is that, since $S$ is convex, $$ \Big\|\frac{y+y'}2-x\Big\|\geq\|y-x\|. $$ Then \begin{align} \|y-y'\|^2 &=\|y-x-(y'-x)\|^2=2\|y-x\|^2+2\|y'-x\|^2-\|y+y'-2x\|^2\\[0.3cm] &=4\|y-x\|^2-4\Big\|\frac{y+y'}2-x\Big\|^2\\[0.3cm] &\leq4\|y-x\|^2-4\|y-x\|^2=0. \end{align} So $y'=y$.