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
Yes, a contraction has a unique fixed point. Proof: Let $d(\cdot,\cdot)$ be our distance function (which is absolute value on the real numbers, but works in more general settings).
Assume $\phi$ is a contraction and $a,b$ are both fixed points. Let $0<k<1$ be our contraction constant. Then $d(g(a),g(b))<k\cdot d(a,b)$, since it's a contraction, but $g(a)=a,g(b)=b$, so $d(g(a),g(b))=d(a,b)$, hence $d(a,b)<k\cdot d(a,b)$, which is only possible if $d(a,b)=0$, hence $a=b$.