Show that $c_0$ is a Banach space with the norm $\rVert \cdot \lVert_\infty$

Suppose $x^k \in c_0$ and $x^k \to x$. Let $\epsilon>0$ and pick $N$ such that $\|x^k-x\|_\infty < {1 \over 2 } \epsilon$ for $k \ge N$. Since $x^N \in c_0$, there is some $N'$ such that $|x_i^N| < {1 \over 2 } \epsilon$ for $i \ge N'$. Then $|x_i| \le |x_i^N| +|x_i-x_i^N| \le |x_i^N| +\|x-x^N\|_\infty <\epsilon$. Hence $x_i \to 0$ and so $x \in c_0$.

Hence $c_0$ is a closed subspace of $l_\infty$.

It follows that $c_0$ is Banach since $l_\infty$ is Banach (any Cauchy sequence in $c_0$ is Cauchy in $l_\infty$ hence converges to some point an closedness shows that this point lies in $c_0$).

We note that $$ |y_k| \leq |y_k - x^n_k| + |x^n_k| = \lim_{m→∞}|x^m_k - x^n_k| + |x^n_k| \leq\lim_{m→∞} ‖x^m - x^n‖ + |x^n_k|$$ which holds for arbitrary $n$. (the superscript is which sequence, the subscript is the index of the sequence) As $x^n∈ c_0$, choose $K_n$ large such that $|x^n_k |<ε$ for $k>K_n$.

Pick $N(ε)$ large such that $\|x^m - x^n||<ε $ for $n,m>N$. Taking limits($\star$) gives $$ \lim_{m→∞}‖x^m - x^n‖ \leq ε \quad ∀ n \geq N$$

So putting the two bounds together: With $ε>0$ given, there is $K = K_{N(ε)}$ such that $$ |y_k| \leq \lim_{m→∞} ‖x^m - x^{N(ε)}‖ + |x^{N(ε)}_k| \leq 2ε $$ for all $k > K$.

The step marked $(\star)$ is easy to see if you notationally suppress the $n$: $$ a_m < ε ∀ m > N \implies \lim a_m \leq ε $$ as (you have shown) this limit will exist.