Let $A \ne \emptyset$ a subset of $\mathbb{R}$, let $f_n:A\to\mathbb{R}$ be a sequence of functions and let $f_n \to f$ uniformly in $A$. My book proves the theorem "a sequence of uniformly convergent bounded functions converges to a bounded function" using the uniformly convergence definition given by the Cauchy convergence, meaning $f_n \to f$ uniformly in $A \iff \forall \varepsilon>0 \ \exists N_{\varepsilon} \in \mathbb{N} \ \text{s.t}. \forall x \in A \ \text{if} \ n,m>N_{\varepsilon}\ \text{then} \ |f_n(x)-f_m(x)| < \varepsilon $.
I've tried to prove it in another way, but I'm not sure if this is correct. Here is my attempt: by hypothesis $f_n$ converges to $f$ uniformly in $A$, meaning that for all $\varepsilon>0$ exists $N_{\varepsilon} \in \mathbb{N}$ such that for all $x \in A$, $n>N_{\varepsilon} \implies |f_n(x)-f(x)|<\varepsilon$. This mean that $-\varepsilon+f_n(x) < f(x) < f_n(x)+\varepsilon$. By hypothesis $f_n$ is bounded in $A$, meaning that there exists $M \in \mathbb{R}$ such that $-M \leq f_n(x) \leq M$ for all $n\in\mathbb{N}$. So it is $-\varepsilon-M \leq -\varepsilon+f_n(x) < f(x) < f_n(x)+\varepsilon<M+\varepsilon$ and this implies that $-\varepsilon-M < f(x) < \varepsilon+M$, and since $\varepsilon>0$ is arbitrary it follows that $-M \leq f(x) \leq M$, so $f$ is bounded.
I suspect that this is wrong because I don't see where I used the uniform convergence, indeed I believe that when I estimate using $-M \leq f_n(x) \leq M$ it isn't correct because the boundedness of $f_n$ is referred to "for all $n\in\mathbb{N}$" and I don't have information if it is valid "for all $x\in A$" too, so to do that I should have something like a "uniform boundedness", meaning I should have that there exists $M \in \mathbb{R}$ such that for all $n \in \mathbb{N}$ and for all $x \in A$ it is $-M \leq f_n(x) \leq M$. It seems like that in my proof the constant $M$ depends on $x$ and this is not ok, but I'm not sure if it is the problem. If it is wrong, can someone explain me why? Thanks.
The main issue in what you did is when you say "by hypothesis $f_n$ is bounded in $A$, meaning that there exists $M \in \mathbb{R}$ such that $-M \leq f_n(x) \leq M$ for all $n\in\mathbb{N}$."
The problem is that $M$ depends on $n$ (not on $x$). To get rid of the issue, you can proceed as follows.
As $\{f_n\}$ converges uniformly to $f$ on $A$, it exists $N \in \mathbb N$ such that $\vert f_n(x) - f(x) \vert \le 1$ for all $x \in A$. Therefore
$$\vert f(x) \vert \le 1 + \vert f_N(x) \vert$$ for $x \in A$. As $f_N$ is supposed to be bounded (we speak of only one map here), it exists $M \gt 0$ such that $\vert f_N(x) \vert \lt M$ for all $x \in A$ and
$$\vert f(x) \vert \le 1 + \vert f_N(x) \vert \le 1 + M$$ for all $x \in A$. Proving as desired that $f$ is bounded on $A$.