= Solution
A <Cauchy sequence> $(x_n)$ in a metric space $(M,d)$ satisfies: for every $\varepsilon>0$ there is $N$ such that $d(x_m,x_n)<\varepsilon$ whenever $m,n\geq N$. A <complete metric space> is one in which every Cauchy <sequence> converges to a point of the space.
Every Cauchy <sequence> is bounded. Choose $N$ such that $d(x_n,x_N)<1$ for $n\geq N$, and put
$$
R=1+\max_{1\leq j<N}d(x_j,x_N).
$$
Then every term lies in the ball $B(x_N,R)$.
Now suppose $M$ is complete and $(F_n)$ is a decreasing <sequence> of nonempty closed sets with $\operatorname{diam}F_n\to0$. Choose $x_n\in F_n$. Given $\varepsilon>0$, choose $N$ with $\operatorname{diam}F_N<\varepsilon$. For $m,n\geq N$, both points lie in $F_N$, so $d(x_m,x_n)<\varepsilon$. Completeness gives $x_n\to x\in M$. For each fixed $N$, the tail lies in the closed set $F_N$, hence $x\in F_N$. Therefore
$$
x\in\bigcap_{n=1}^{\infty}F_n.
$$
Conversely, assume the nested-set property and let $(x_n)$ be Cauchy. Define
$$
F_n=\overline{\{x_m:m\geq n\}}.
$$
These sets are nonempty, closed, and decreasing. The Cauchy property implies $\operatorname{diam}F_n\to0$; taking a closure does not change the diameter. Choose $x\in\bigcap_nF_n$. Since $x_n,x\in F_n$,
$$
d(x_n,x)\leq\operatorname{diam}F_n\longrightarrow0.
$$
Thus every Cauchy <sequence> converges and $M$ is complete. This proves the <Cantor intersection theorem> characterization.
The <contraction mapping theorem> states that a contraction of a nonempty complete metric space has a unique fixed point.
For each $\lambda\in\Lambda$, the map $T_\lambda(x)=T(\lambda,x)$ is a contraction with the common constant $k$, so it has a unique fixed point $x^*(\lambda)$. This defines the required unique <function>. Fix $\lambda_0$. The fixed-point identities and the triangle inequality give
$$
\begin{aligned}
d(x^*(\lambda),x^*(\lambda_0))
&\leq d(T(\lambda,x^*(\lambda)),T(\lambda,x^*(\lambda_0)))\\
&\quad+d(T(\lambda,x^*(\lambda_0)),T(\lambda_0,x^*(\lambda_0)))\\
&\leq k\,d(x^*(\lambda),x^*(\lambda_0))
+d(T(\lambda,x^*(\lambda_0)),T(\lambda_0,x^*(\lambda_0))).
\end{aligned}
$$
Consequently
$$
d(x^*(\lambda),x^*(\lambda_0))
\leq\frac{d(T(\lambda,x^*(\lambda_0)),T(\lambda_0,x^*(\lambda_0)))}{1-k}.
$$
The numerator tends to zero as $\lambda\to\lambda_0$ by the assumed continuity for the fixed point $x^*(\lambda_0)$. Hence $x^*$ is continuous, an instance of <continuous dependence of the fixed point of a uniform contraction>.
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