Problem Statement:
Let $\{f_n\}$ converge uniformly on a set $E$. Suppose that each $f_n$ is bounded. Prove that there exists an $M > 0$ s.t. $|f_n(x)| \leq M$ for all $x \in E$ and all $n = 1,2, ...$.
Proof:
If each $f_n$ is bounded by some $M_n > 0$ then can't we just take $M = \text{max}\{M_n\}$?
I don't see what's so difficult here. Am I not using the fact that $\{f_n\}$ converges uniformly?
On
You haven't proven that $M<\infty$ - an infinite set need not have a maximum. As commented, if for example $M_n = n$, then indeed each $M_n$ is finite, but there is no finite $M$ such that $M_n\leq M$ for all $n$.
By uniform convergence and the triangle inequality, $$|f(x)| \leq |f_n(x)-f(x)| + |f_n(x)|< \varepsilon + M_n < \infty$$
Note that the right hand side is independent of $x$ so we are done.
Hint: if $f_n \to f$ uniformly, then $$\|f_n\|_\infty \leq 1+\|f\|_\infty$$ for every $n \gg 1$. By a standard argument, there exists a constant $C>0$ such that $$\|f_n\|_\infty \leq C + 1 + \|f\|_\infty$$ for every $n$.
Of course I denoted $\|f\|_\infty = \sup_{x \in E} |f(x)|$ and used the fact that $f_n \to f$ uniformly on $E$ means $\lim_{n \to +\infty} \|f_n-f\|_\infty =0$.