Every finite-dimensional normed space is Banach

By [L1], it is enough to prove that ${\mathbb{K}}^n$ is complete for the coordinate ${\mathrm{\ell }}^1$ norm, because a bounded bijection with bounded inverse preserves Cauchy sequences and their limits.

In the real case $\mathbb{K}=\mathbb{R}$, if $n=0$ then ${\mathbb{R}}^0=\{0\}$ is complete trivially. If $n\ge 1$, [L2] applies directly to the ${\mathrm{\ell }}^1$ norm on ${\mathbb{R}}^n$, so ${\mathbb{R}}^n$ is complete.

In the complex case, if $n=0$ the coordinate space is a singleton and complete. If $n\ge 1$, let $\mathrm{\Psi }:{\mathbb{C}}^n\to {\mathbb{R}}^{2n}$ send each coordinate to its real and imaginary parts. By [L3] it is a real-linear bijection, and $Q(u):=\mathrm{\parallel }{\mathrm{\Psi }}^{-1}u{\mathrm{\parallel }}_1$ is a real norm on ${\mathbb{R}}^{2n}$. It is complete by [L2]. Since $Q(\mathrm{\Psi }a-\mathrm{\Psi }b)=\mathrm{\parallel }a-b{\mathrm{\parallel }}_1$, the bijection $\mathrm{\Psi }$ transports Cauchy sequences and limits exactly; thus ${\mathbb{C}}^n$ is complete in its complex coordinate norm metric.

Let $(x_m)$ be a Cauchy sequence in $X$, and write $a_m:=T^{-1}{x}_m$. Since $T^{-1}$ is bounded, $(a_m)$ is Cauchy in ${\mathbb{K}}^n$; by steps 1.2 and 1.3 it converges to some $a\in {\mathbb{K}}^n$. Since $T$ is bounded, $x_m=T(a_m)\to T(a)$ in $X$. Thus every Cauchy sequence in $X$ converges in $X$.

By [L4], step 2.1 says exactly that $X$ is Banach.