Proposition : A convergent sequence is bounded.
Proof : Let $\{x_{n}\}$ be a sequence that converges to $L$. Then by definition, $|x_{n} - L| < \epsilon$ $\forall \epsilon > 0$. Suppose $\epsilon = 1$. Then we have $|x_{n} - L| < 1$. But $|x_{n}| = |x_{n} - L + L| \leq |x_{n} - L| + |L| < 1 + |L|$ so $|x_{n}| < 1 + |L|$.
Let $M = \max(|a_{1}|,|a_{2}|,...,|a_{N}|, 1+|L|)$. Then $x_{n} \leq M$ $\forall n$.
I sort of understand the last sentence in this proof, but then I don't. It is mostly how $1 + |L|$ does not necessarily bound the whole sequence. Could a concrete example be provided where $1+ | L|$ does not bound the whole sequence?
You just have by definition of the limit that there exists an $N\in\Bbb N$ s.t. for $n\ge N$ it holds $|x_n - L| < 1$. But what's for $n<N$? There it don't need to hold.
E.g. Take $x_n = \frac{100}{n}$, then $L=0$ but $|x_n - L| < 1$ hold's only for $n\ge 101$ but $|x_n - L| \ge 1$ for $n \le 100$.
So you have to choose $M = \max\{a_1,\ldots,a_n,1\} = 100$ as the boundary of $x_n$