Prove that $f_n$ is uniformly bounded on $E$ and $f$ is a bounded function on $E$ I have the solution for the following problem, but I don't understand most of it. 
Question: A sequence of functions $f_n$ is said to be uniformly bounded on a set $E$ iff there exists $M>0$ such that $|f_n(x)\le M$ for all $x\in E$ and all $n\in \mathbb{N}$.
Suppose that for each $n\in\mathbb{N}$, $f_n:E\to\mathbb{R}$ is bounded. If $f_n\to f$ uniformly on $E$, as $n\to\infty$, prove that ${f_n}$ is uniformly bounded on $E$ and $f$ is a bounded function on $E$.
Solution:
Choose $N$ so large that $|f(x)-f_n(x)|<1$  for all $x\in E$ and $n\ge N$.
Why does it have to be less than $1$?
Set $M:=\sup_{x\in E}|f_N(x)|$ and observe by the Triangle Inequality that
$$|f(x)|\le|f(x)-f_N(x)|+|f_N(x)|<1+M \text{ for all }x\in E\text{.}$$
Therefore, $|f_n(x)|\le|f(x)|+1\le(1+M)+1=2+M$ for all $n\ge N$ and $x\in E$, i.e., $\{f_n\}_{n\ge N}$ is uniformly bounded on $E$.
I understand this, but not what is below. What exactly is it trying to show? And how does that prove that $f$ is a bounded function on $E$?
In particular,
$$|f_n(x)\le\hat{M}:=\max\{2+M,\sup_{x\in[a,b]}|f_1(x)|,...,\sup_{x\in[a,b]}| f_{N-1}(x)|\}<\infty$$
for all $n\in\mathbb{N}$ and $x\in E$.

Answers, in order:


*

*the choice of $1$ was arbitrary; this would work for any $\epsilon > 0$, so we might as well choose $\epsilon = 1$.  As Daniel Fischer notes, $\epsilon = \pi$ and $\epsilon = 10^{10^{1000}}$ would work just as well, for the purposes of the proof.

*in the above, we've shown that $|f(x)| \leq 1 + M$, so we know that $f$ is bounded.  We also know that for any $n \geq N$, $f_n$ is also bounded by $2+M$ so that $\{f_n\}_{n \geq N}$ is uniformly bounded.  What remains to be shown is that the first elements $\{f_1,\dots,f_{N-1}\}$ have a common bound, so that the entire sequence $\{f_n\}_{n \geq 1}$ can be shown to be bounded.  In the last line, we establish that the entire sequence is in fact uniformly bounded.

