Prove convergence in $\Bbb R^m$ My attempt:
First, I have this information:
First, for any $\epsilon_{1}>0$, $\exists N_{1}∈\mathbb{N}$ such that $\left \| x_{n}-a \right \|<\epsilon_{1}, \quad\forall n\geq N_{1}$.

It’s basically correct, but it could be organized a bit better. In particular, you don’t need different $\epsilon$s and $N$s. Here’s one possibility:

By hypothesis there is a $c>0$ such that $|\lambda_n|<c$ for each $n\in\Bbb N$. Let $\epsilon>0$. Since $\lim_{n\to\infty}x_n=0$, there is an $N\in\Bbb N$ such that $\|x_n\|=\|x_n-0\|<\frac{\epsilon}c$ whenever $n\ge N$. But then $\|\lambda_nx_n\|=|\lambda_n|\|x_n\|<c\cdot\frac{\epsilon}c=\epsilon$ whenever $n\ge N$, so $\lim_{n\to\infty}\lambda_nx_n=0$.

