Metric in Hilbert Space defined as $d := d(x, y) = \inf_{\lambda} \|\ x + \lambda y \|\ $, where $\|\ y \|\ = 1 $ 
Question: Let $d := d(x, [[y]]) = \inf_{\lambda} \|\ x + \lambda y \|\ $, where $y$ is a unit vector; and $[[y]]$ denotes the span of vectors (with norm of 1); show that:
(a) $d = \|\ x + \lambda_0 y \|\ $ for some $\lambda_0$ ,
(b) $| \langle x,y \rangle |^{2} = \|\ x \|^{2} - d^2 $ , and
(c) $y \perp (x - \lambda_0 y )$

It's also worth mentioning that we're working in a Hilbert Space. Also, I'm given a hint: substitute $\lambda = \alpha + i \beta$, then find the minimum by differentiating in $\alpha$ , $\beta$ to get $\lambda = - \overline{ \langle x , y \rangle } $. But when I tried to use it nothing surfaced for me.
So here are my thoughts so far:
Since any vector $x$ can be decomposed uniquely into two parts, one in the direction of $y$, and the other perpendicular to it:
$$x = \lambda y + (x - \lambda y) \hspace{1cm} \text{with} \space\ \langle x - \lambda y , y \rangle = 0$$
which shows part (c), (feels wrong) but any Hilbert Space has the representation $H = M \bigoplus M^{\perp} $ where $M$ is a closed subspace of $H$. Anyways, this yields $\lambda = \dfrac{ \langle y , x \rangle }{ \langle y , y \rangle } $, but in our case $\|\ y \|\ = 1$, so we have $\lambda = \langle y , x \rangle = \overline{ \langle x , y \rangle } $. Applying Pythagoras' Theorem we have that:
$$ \|\ x \|^2 = \|\ \lambda y \|^2 + \|\ x - \lambda y \|^2 $$
Then, substituting in what the hint indicated I should of got (which I didn't); I would have $\lambda = - \overline{ \langle x , y \rangle }$ would result in:
$$\|\ x \|^{2} = | \lambda |^{2} \|\ y \|^{2} + \|\ x + \lambda y \|^{2} \iff | \langle x,y \rangle |^{2} = \|\ x \|^{2} - d^2 $$
showing part (b).
 A: Note that $[y]=\overline {[y]}$ because, using the fact that the norm is continuous, we have 
$\tag 1\lambda_ny\to z\in \overline {[y]}\Rightarrow |\lambda_n|\to \left \| z \right \|$  
Thus, $\left \| z-\left \| z \right \|\cdot y \right \|\le \left \| z-\lambda_ny \right \|+\left \| \lambda_ny- \left \| z \right \|\cdot y\right \|$ and by $(1)$, each term can be made as small as we like and this implies that $z=\left \| z \right \|\cdot y$.
Now the Projection Theorem gives us a vector $x'$ and a scalar $\lambda_0$ such that $x=x'+\lambda_0y,\ $ with 
$\tag2 \left \| x-\lambda_0 y \right \|\  \text {is minimal}.$
and  $\tag3(x',y)=0.$ 
$(2)$ says $\lambda_0=d,\ $ which is $(a)$. $(3)$ gives $(c)$ almost immediately, and finally, an application of the  Pythagorean Theorem gives $(b)$.
Remark: to do $(a)$ using the hint, since $x$ is fixed, we can choose a scalar $\beta $  such that $(\beta y,x)$ is real. So wlog assume $(y,x)$ is real.  Then, $\phi$ defined by $\phi(\lambda)=\left \| x+\lambda y \right \|^2=\left \| z \right \|^{2}+2\lambda (y,x)+\lambda^{2}\left \| y \right \|^{2}$ is a quadratic in $\lambda$ and has a minimum at $\lambda =-(y,x).$
