From
$$d(x_{n+1},x_n)=d(g(g(x_{n-1})),g(x_{n-1}))\le \lambda d(g(x_{n-1}), x_{n-1})=d(x_n,x_{n-1})$$
we get after $n$ applications of that inequality
$$d(x_{n+1},x_n)\le \lambda d(x_n,x_{n-1}) \le \lambda^2 d(x_{n-1},x_{n-2}) \le \cdots\le \lambda^n d(x_1,x_0)\tag{1}$$
Now we want to show that $(x_n)_n$ is a Cauchy sequence. So let $\epsilon>0$.
We assume $x_1\not=x_0$ (otherwise $x_0$ is already a fixed point). Set $c=d(x_1,x_0)>0$.
Since $0<\lambda<1$, the sum $\sum_{n=0}^\infty \lambda^n$ converges (to $1/(1-\lambda)$). Therefore we can pick $N$ large enough such that
$$\sum_{k=n}^\infty \lambda^k<\frac{\epsilon}{c}$$
for all $n\ge N$.
Then for $m>n\ge N$ we have by the triangle inequality
$$d(x_m,x_n)\le \sum_{k=n}^{m-1} d(x_{k},x_{k+1})$$
Applying $(1)$ we obtain
$$d(x_m,x_n)\le c\sum_{k=n}^{m-1} \lambda^k\le c\sum_{k=n}^\infty \lambda^k<c\cdot\frac{\epsilon}{c}=\epsilon$$
So $(x_n)_n$ really is a Cauchy sequence. Since $(X,d)$ is complete, it converges to a limit $x\in X$.
By the equation $x_{n+1}=g(x_n)$, the limit satisfies $x=g(x)$, so it is a fixed point.
Uniqueness is trivial, let $y$ be another fixed point of $g$. Then
$$d(x,y)=d(g(x),g(y))\le \lambda d(x,y)$$
If now $x\not=y$, then $d(x,y)>0$, so we can divide by $d(x,y)$ to obtain $\lambda\ge 1$, a contradiction. Therefore, $x=y$.
Best Answer
Because that's what the theorem states, and therefore, that's what the proof should conclude with.
What do you mean by "original proof"? There is only one proof and it can be divided into the following parts:
Now, sure, alternatively, the authors could cite point number 2 above as a separate lemma, and prove it before proving the main theorem, but I'd say it's cleaner this way, for two reasons:
Edit:
It is, of course, also possible to prove point 3 directly from 1 and from the fact that $T$ is continuous. However, this would most probably lead to a proof that is harder to follow, and possibly even longer than original.
But the mere fact that you can prove something without a particular sub-lemma does not mean that that particular lemma is "useless" for the proof. Technically, every proof in mathematics can be trimmed down to base principles, but at a great cost of readability.