Proving that the set of fixed points of a nonexpansive operator on a bounded closed convex subset of a Hilbert space is convex I'm struggling to understand the proof of Lemma $3.1.$ from the book Iterative Approximations of Fixed Points by Vasile Berinde, which I'll transcribe here

Lemma 3.1. Let $C$ be a bounded closed convex subset of a Hilbert space $H$ and $T:C \rightarrow C$ be a nonexpensive and demicompact
operator. Then the set $F_T$ of fixed points of $T$ is a nonempty
convex set.

The proof starts by using a previous result (Theorem $3.1.$) in which he shows that $F_T$ is nonempty, and this is fine. I'm struggling in verifying the convexity of $F_T$. The original argument starts by letting $x,y \in F_T, \lambda \in [0,1]$, and writing $u_\lambda = (1-\lambda)x+\lambda y$. Then observe that
$$ \|T(u_\lambda) - x\| = \|T(u_\lambda) - T(x)\| \le \|u_\lambda - x\| $$
$$ \|T(u_\lambda) - y\| = \|T(u_\lambda) - T(y)\| \le \|u_\lambda - y\|. $$
And it follows that
\begin{align}
\| x- y \| &= \| x - T(u_\lambda) + T(u_\lambda) - y\| \\
&\le \|x-T(u_\lambda)\| + \| T(u_\lambda) - y\| \\
&\le \| u_\lambda -x\| + \| u_\lambda - y\| \\
&= \lambda\|x-y\| + (1-\lambda)\|x-y\| \\
&= \| x-y\|, 
\end{align}
and this is fine too. But at this point, the author claims that this last fact guarantees the existence of some $a\ge 0$ and $b \le 1$ such that
$$ x-T(u_\lambda) = a(x-u_\lambda) $$
$$ y-T(u_\lambda) = b(y-u_\lambda), $$
from which it follows that $T(u_\lambda)=u_\lambda \in F_T$.
I am not understanding this very last part of the argument, in which he invokes the existence of constants $a$ and $b$. My first guess was that he was somehow using the strict convexity of $H$, but this would only provide $c,d>0$ such that
\begin{align}
x-T(u_\lambda) &= c (T(u_\lambda) - y) \\
x-u_\lambda &= d (u_\lambda-y).
\end{align}
What am I missing here?
 A: From your inequalities you get
$$
\|T(u_\lambda) - x\| =  \|u_\lambda - x\|,\qquad\qquad
\|T(u_\lambda) - y\| = \|u_\lambda - y\|,
$$
Then you deduce (since you can discard the cases $u_\lambda=x $ and $u_\lambda=y $ from the start) that $c=d$.  Thus
\begin{align}\tag1
x-T(u_\lambda) &= c (T(u_\lambda) - y) \\ \tag2
x-u_\lambda &= c (u_\lambda-y).
\end{align}
Subtracting $(2)$ from $(1)$,
$$
u_\lambda-Tu_\lambda=c(Tu_\lambda-u_\lambda).
$$
Looking at the norms, $c=1$ and then $Tu_\lambda=u_\lambda$.
A: The equality
$$\| x - T(u_\lambda) + T(u_\lambda) - y\| \\
=\|x-T(u_\lambda)\| + \| T(u_\lambda) - y\|$$
implies that $T(u_\lambda)$ belongs to the interval connecting $x$ and $y.$ Also $u_\lambda$ belongs to that interval (by definition). Therefore $$x-T(u_\lambda)=a(x-u_\lambda),\ a\ge 0 \\ y-T(u_\lambda)=b(y-u_\lambda),\ b\ge 0$$ Since $T$ is nonexpansive we get $a,b\le 1.$
We have two points $T(u_\lambda)$ and $u_\lambda$ in the segment connecting $x$ and $y.$ The distances from $T(u_\lambda)$ to $x$ and $y$ are shorter or equal to the distances of $u_\lambda$ to $x$ and $y,$ respectively. This is possible only when $T(u_\lambda)=u_\lambda.$
