Proof that convergent sequence in $\Bbb R$ is bounded. Convergent sequence in $\Bbb R$ is bounded. Proof: 

In the definition of a convergent sequence: 
$$(\forall \varepsilon>0), (\exists n_\varepsilon\in\Bbb N), (\forall n\in\Bbb N), ((n>n_\varepsilon)\Rightarrow(|a_n-a|<\varepsilon))$$ 
  let $\varepsilon=1$, then there exists $n_\varepsilon\in\Bbb N$ such
  that $(n>n_\varepsilon)\Rightarrow(|a_n-a|<1$). Now for
  $n>n_\varepsilon$ we have $|a_n|\leq|a_n-a|+|a|\leq 1+|a|$. Let
  $M=\max\{|a_1|,...,|a_{n_\varepsilon}|,1+|a|\}$ Then $\forall n\in\Bbb
> N, \ |a_n|\leq M$, i.e. the sequence is bounded. 

What's the idea behind this proof? I understand the first part, but then when they define $M$ I'm lost. I don't see how is that related to the first part of the proof. 

The idea behind this proof is to separate the values of the sequence into 2 parts : the ones that are close to the limit (within distance 1 of it) and the ones that are "far" from the limit.
By definition of convergent sequences, those last ones are in finite number, so the set of their absolute values is bounded, by a real $M$. And the set of the others is bounded by $|l|+1$.
So the whole set of absolute values is bounded by $\max(|l|+1,M)$.
