Let $(x_n)_{n \in \mathbb{N}}$ be a convergent sequence, lets say $(x_n)_{n \in \mathbb{N}}$ converges to $x$. By definition, for every $\epsilon > 0$ there exist $N \in \mathbb{N}$ such that $||x_n -x||< \epsilon$ for every $n \geq N$. From here it is clear that for every $n \geq N$ the sequence is bounded because $$ ||x_n|| = ||x_n-x +x|| \leq ||x_n-x||+||x|| < \epsilon+||x|| $$
For $n<N$ we can define $M = \max\{\epsilon+||x||,||x_1||,\dots,||x_{N-1}|| \}$, then $||x_n||<M$.
I have trouble with the last part, lets say we take $a\in \mathbb{N}$, for example, $a=3,$ then the sequence $(\frac{1}{|n-3|})_{n \in \mathbb{N}}$ converges to $0$ as $n \rightarrow \infty$. By definition, for every $\epsilon >0$ there exist $N\in\mathbb{N}$ such that $|\frac{1}{|n-a|}|<\epsilon $ for every $n\geq N$. However, for $n < N$, if we define $M = \max\{\epsilon, \frac{1}{|1-3|,}, \frac{1}{|2-3|}, \frac{1}{|3-3|}, \dots, \frac{1}{|(N-1)-3|}\}$, it is clear that $\frac{1}{|3-3|}$ is undefined, does this mean we just exclude that possibility for the maximum?
