Prove that $(\ell ^∞,||·||_∞)$ is a Banach space using the following steps.
Let $(x_n)_{n∈\mathbb N}$ be a Cauchy sequence in $(\ell ^∞,||·||_∞)$. For $n > 1$, let $x_n = (a_{1,n},a_{2,n},...,a_{k,n},...)$.
(a) Prove that for every fixed $k \geq 1$, the sequence of real numbers $(a_{k,n})_{n>1}$ is a Cauchy sequence in $\mathbb R$. Therefore conclude that $(a_{k,n})_{n>1}$ converges to a limit $y_k ∈ \mathbb R$.
Steps (b), (c), (d), I didn't write yet.
Let $\varepsilon /2 >0$. Then $\exists N $ such that $n,m > N$, we have:
$$|| a_{k,n} - a_{k,m}||_{\infty} = \sup _{k \geq 1} | a_{k,n} - a_{k,m}| \leq\sup _{k \geq 1} | a_{k,n} |+\sup _{k \geq 1} | - a_{k,m}| = \sup _{k \geq 1} | a_{k,n} |+\sup _{k \geq 1} | a_{k,m}| $$
I kind of guessed all of this so far so it might be wrong... I don't really know what to do. Please help.
For $y\in \ell^\infty$ , $y= (y_1,y_2,\dots)$, and $k\in \mathbb N$ we have that $$|y_k|\leq \sup_{k \in \mathbb N}|y_k|= \| y\|_\infty.$$
If $(x_n)_{n \in \mathbb N} = (a_{1,n}, a_{2,n}, \dots)$ is a Cauchy sequence, then we have by definition that for every $\epsilon > 0$ there exists some $N \in \mathbb N$ such that we can conclude from $n,m > N$ that $\|x_n -x_m\|< \epsilon$.
Let $k\in \mathbb N$. For $\epsilon > 0$, we choose $N$ like above and insert $y = x_n - x_m$ in the equation. We get that for $n,m > N$ $|a_{k,n} - a_{k,m}|<\epsilon$. This means that the sequence $(a_{k,1}, a_{k,2}, \dots$ (not that $k$ is fixed) is a real sequence which is Chauchy as well, i.e. it converges to some $a_{k}\in \mathbb R$.