Solution: Let $ \left( f_n \right) $ be a sequence of functions which converges uniformly on a domain $D$. Further assume that each $f_n$ is bounded. That means, for every $n \in \mathbb{N} $ there exists $M_n \in \mathbb{R} $ such that for any $x \in D$, we have \[ \left\vert f_n(x) \right\vert \leq M_n. \] We need to show that the set \[ \mathcal{F} = \left\{ f_n: n \in \mathbb{N} \right\} \] is uniformly bounded. Let $\epsilon = 1$. since the sequence of function converges uniformly on $D$, they are also uniformly Cauchy. This means, we can find $n_0 \in \mathbb{N} $ such that for every $x \in D$ and for $m,n \geq n_0$, we have \[ \left\vert f_n (x) - f_m(x) \right\vert < 1. \] This implies, for $n \geq n_0$, \begin{align*} \left\vert \vert f_n(x) \vert - \vert f_{n_0}(x) \vert \right\vert \leq \left\vert f_n(x) - f_{n_0}(x) \right\vert < 1 \\ \implies \left\vert f_n(x) \right\vert \leq 1 + \vert f_{n_0}(x) \vert \leq 1 + M_{n_0}. \end{align*} Let \[ M \coloneqq \max \left\{ M_{n_0} + 1, M_1, M_2, \dots, M_{n_0 - 1} \right\} . \] Then for any $n \in \mathbb{N} $, and for any $x \in D$, we have $\vert f_n(x) \vert < M$ and hence $\mathcal{F} $ is uniformly bounded.