set $U_{2}$ to get $x_{2} \in B\left(x_{1}, r_{1}\right) \cap U_{2}$. Again, we can find $r_{2}$ such that $0<r_{2}<2^{-2}$ and $B\left[x_{2}, r_{2}\right] \subset B\left(x_{1}, r_{1}\right) \cap U_{2}$. Proceeding this way, for each $n \in \mathbb{N}$, we get $x_{n} \in X$ and an $r_{n}$ with the properties
$$
B\left[x_{n}, r_{n}\right] \subset B\left(x_{n-1}, r_{n-1}\right) \cap U_{n} \text { and } 0<r_{n}<2^{-n}
$$
Clearly, the sequence $\left(x_{n}\right)$ is Cauchy: if $m \leq n$,
$$
d\left(x_{m}, x_{n}\right) \leq d\left(x_{n}, x_{n-1}\right)+\cdots+d\left(x_{m+1}, x_{m}\right) \leq \sum_{k=m}^{n} 2^{-k}
$$
Since $\sum_{k} 2^{-k}$ is convergent, it follows that $\left(x_{n}\right)$ is Cauchy.
MATH4061 COURSE NOTES ：
We shall sketch the argument. Let $m, M$ be such that
$$
m \leq \frac{\partial f}{\partial y}(x, y) \leq M, \text { for }(x, y) \in D
$$
Let $X:=\left(C[a, b], |_{\infty}\right)$. Consider
$$
T(\varphi)(x):=\varphi(x)-\frac{1}{M} f(x, \varphi(x)), \text { for } x \in[a, b]
$$
One shows that $T$ is a contraction by an obvious use of the mean value theorem.