Strict convexity implies there exists $\theta \in (0,1)$ such that $Tz=\theta u+(1-\theta)v$. 
Let $X$ be a strictly convex Banach space and $Y$ be a closed, convex subset of $X$. Let $T:Y\to Y$ be a mapping satisfying $\lVert u-Tv\rVert \le \lVert u-v\rVert+\lVert u-Tu\rVert$. Show that ${\rm Fix}(T)$ is convex set, where ${\rm Fix}(T)$ denotes the set of all fixed points of $T$.

Proof: Let $\eta \in (0,1)$ and $u,v\in {\rm Fix}(T)$ with $u\ne v$ and put $z:=\eta u+(1-\eta)v$. Then we have,
$\lVert Tu-Tz\rVert=\lVert u-Tz\rVert\le \lVert u-z\rVert+\lVert u-Tu\rVert=\lVert u-z\rVert.$
Similarly, $\lVert Tv-Tz\rVert\le \lVert v-z\rVert.$
Again, $\lVert u-v\rVert\le \lVert u-Tz\rVert+\lVert v-Tz\rVert=\lVert Tu-Tz\rVert+\lVert Tv-Tz\rVert\le\lVert u-z\rVert+\lVert v-z\rVert=\lVert u-v\rVert$
Since, $X$ is strictly convex, there exists $\theta \in (0,1)$ such that $Tz=\theta u+(1-\theta)v$.
Then we have, $(1-\theta)\lVert u-v\rVert=\lVert Tu-Tz\rVert\le \lVert u-z\rVert=(1-\eta)\lVert u-v\rVert$ and
$\theta \lVert u-v\rVert=\lVert Tv-Tz\rVert\le \lVert v-z\rVert=\eta \lVert u-v\rVert$. Therefore, $1-\theta \le 1-\eta$ and $\theta \le \eta$ implies $\theta =\eta$. Then $z=Tz$. That is $z\in {\rm Fix}(T)$.
Question: In this proof I'm auable to understand the line in bold text. Can anyone help me to understand how strict convexity implies that $Tz$ is the convex combination of $u$ and $v$ ?
 A: I am not sure if in the critical step you can directly show the existence of such a $\theta$ in $(0,1)$ but at least in $[0,1]$ including 0 and 1. But the remaining steps remain valid even in these cases. So the proof is fine.
Hence let us show that the strict convexity implies the existence of $\theta\in[0,1]$ with $Tz=\theta u + (1-\theta)v$.
If $Tz$ equals u or v, then we can choose $\theta=1$ or $\theta=0$, respectively. So let us assume $Tz\ne u$ and $Tz\ne v$ for the remaining cases. From the bold line above the critical one we get $$||u-Tz|| + ||Tz-v|| = ||u-v||.$$ Set $a:=u-Tz$ and $b:=Tz-v$ this means $||a||+||b||=||a+b||$ with $a,b\ne0$. Either you already know that strict convexity of $X$ implies that $a$ and $b$ are on the same line. Or you can proceed as follows: Setting $\alpha:=\frac{||a||}{||a||+||b||}\in(0,1)$, $a':= a/||a||$ and $b':=b/||b||$ we have
$$ \alpha a' + (1-\alpha)b' = \frac {a+b} {||a||+||b||} = \frac{a+b}{||a+b||}. $$
Hence, $||a'||=1$, $||b'||=1$ and $||\alpha a' + (1-\alpha)b'||=1$. If $a'\ne b'$ then the strict convexity of $X$ implies a contradiction. Hence $a'=b'$ and therefore there is a $t>0$ with $a=tb$.
But then $t(Tz-v)=u-Tz$ and therefore $$Tz = \frac{1}{1+t} u + \frac{t}{1+t}v.$$ Now let $\theta:=1/(1+t)$ and we are done.
