Step 1: Construct candidate sequence. Fix $x_0 \in X$ and define $x_{n+1} = f(x_n)$ for $n \geq 0$.

Step 2: Show $(x_n)$ is Cauchy. For $n \geq 1$,

$d(x_{n+1}, x_n) = d(f(x_n), f(x_{n-1})) \leq \lambda d(x_n, x_{n-1}) \leq \lambda^2 d(x_{n-1}, x_{n-2}) \leq \cdots \leq \lambda^n d(x_1, x_0).$

For $m > n$, by the triangle inequality,

$d(x_m, x_n) \leq d(x_n, x_{n+1}) + d(x_{n+1}, x_{n+2}) + \cdots + d(x_{m-1}, x_m) \leq \sum_{k=n}^{m-1} \lambda^k d(x_1, x_0) = \lambda^n \frac{1 - \lambda^{m-n}}{1 - \lambda} d(x_1, x_0).$

Since $\lambda < 1$, $\lambda^n \to 0$, so $d(x_m, x_n) \to 0$ as $n \to \infty$. Thus $(x_n)$ is Cauchy.

Step 3: Convergence. By completeness of $X$, there exists $x^* \in X$ such that $x_n \to x^*$.

Step 4: $x^*$ is a fixed point. By continuity of $f$ (a consequence of the Lipschitz condition),

$f(x^*) = f(\lim_{n \to \infty} x_n) = \lim_{n \to \infty} f(x_n) = \lim_{n \to \infty} x_{n+1} = x^*.$

Step 5: Uniqueness. Suppose $f(y) = y$ for some $y \in X$. Then

$d(x^*, y) = d(f(x^*), f(y)) \leq \lambda d(x^*, y).$

Since $0 \leq \lambda < 1$, this implies $(1 - \lambda) d(x^*, y) \leq 0$, so $d(x^*, y) = 0$. Thus $x^* = y$.

Step 6: Error estimate. From Step 2, taking $m \to \infty$,

$d(x_n, x^*) \leq \frac{\lambda^n}{1 - \lambda} d(x_1, x_0).$

This shows the iteration converges exponentially fast.