Here is a way to show this when you consider the field of real numbers and this for all norms.
Consider $E$ a finite dimensional vector space over $\mathbb{R}$ and a basis $(e_1, ..., e_n)$ of $E$. It means that for all $x\in E$ you have $x = \sum_{i=1}^{n} x_{i}e_i $ where the $x_i$ are real numbers.
Now we need to endow E with a norm, first recall that all norms are equivalents in a finite dimensional normed space it will be usefull at some points.
The norm we will consider is $\lVert x\rVert_{\infty} = max_{1\leq i\leq n}\lvert x_i\rvert $. Thus $(E,\lVert\rVert_{\infty})$ is a finite dimensional normed space.
Now consider a Cauchy sequence in $E$, namely $(x^{p})_{p\in\mathbb{N}}$, it means that
$\forall\epsilon>0,\exists N\in\mathbb{N} : n,m\geq N\implies\lvert x_i^{n} -x_i^{m}\rvert\leq\lVert x^{n} - x^{m}\rVert_{\infty} = max_{1\leq i \leq n}\lvert x_{i}^{n} - x_{i}^{m}\rvert\leq\epsilon$
However for $i\in\{1,...,n\}$ we have that $(x_i^{p})$ is a sequence of real numbers and we have shown that it is Cauchy !
Thus for all $i\in\{1,..n\}$ there exists $x_i\in\mathbb{R}$ such that $\forall\epsilon>0, \exists N_{i}\in\mathbb{N} : p\geq N_i\implies \lvert x_i^{p} -x_i\rvert\leq\epsilon$ (since $\mathbb{R}$ is complete)
Now, a candidate we should consider as a limit of $(x^p)_{p\in\mathbb{N}}$ is the vector whose coordinate corresponds to the $x_i$'s that is $x = \sum_{i=1}^{n}x_ie_i$.
Fix $\epsilon>0$ and $N = max_{1\leq i\leq n}(N_i)$, then we have that
$p\geq N\implies max_{1\leq i\leq n}\lvert x_i^p - x_i\rvert = \lVert x^p - x\rVert_{\infty}\leq\epsilon$ which shows the convergence of $(x^p)_{p\in\mathbb{N}}$ to $x\in E$.
Since $E$ is a finite dimensional normed space, all norms on this space are equivalents which means that the convergence is not altered by a change of norm (this is already explained in the comments I think)
I hope this is clear !
Let $\epsilon >0$. There exists $N$ such that $\sup _t |f_n(t)-f_m(t)| +K(f_n-f_n)<\epsilon$ for all $n,m >N$. This implies that $\sup _t |f_n(t)-f_m(t)| <\epsilon$ for all $n,m >N$ and $K(f_n-f_m)<\epsilon$ for all $n,m >N$. $(f_n(t))$ is a Cauchy sequence so $f(t)=\lim f_n(t)$ exists for each $t$. Letting $m \to \infty$ in the inequality $|f_n(t)-f_m(t)| <\epsilon$ we get $|f_n(t)-f(t)| \leq \epsilon$ for $n >N$. Similarly, we can let $m \to \infty$ in $K(f_n-f_m) <\epsilon$ to get $k(f_n-f) \leq \epsilon$ for $ n>N$. Rest of the argument should now be clear.
Best Answer
Hints: Elements of this space are themselves sequences. So a Cauchy sequence is of the type $(x^{n}_1,x^{n}_2,...)$ , $n=1,2...$ with the property that $\sup_i |x^{n}_i-x^{m}_i| \to 0$ as $n,m \to \infty$. When this holds $(x^{1}_i,x^{2}_i,...)$ is a Cauchy sequence of real numbers for each $i$. So $x_i=\lim_{n \to \infty} x^{n}_i$ exists for each $i$. This gives a new sqequence $(x_i)$. Now try to show that this sequence is bounded and the original Cauchy sequence converges to this element.