 Solution

The Baire category theorem in the needed version states that a nonempty complete metric space cannot be a countable union of closed sets with empty interior. To prove it, suppose $F_1,F_2,\dots$ are such sets and start with a nonempty open set $U$. Inductively choose closed balls ${\overline{B}}_j$ of positive radius at most $2^{-j}$, with ${\overline{B}}_1\subset U\setminus F_1$ and ${\overline{B}}_{j+1}\subset B_j\setminus F_{j+1}$. This is possible because each complement is open and dense. The centres form a Cauchy sequence; completeness supplies a limit in every closed ball. It lies in $U$ and outside every $F_j$, proving the theorem.

Here the sequence spaces are over the real scalars, as required by the Euclidean-plane assertion. Suppose an isometry $\phi :{\mathbb{R}}^2\to c_0$ existed. On the compact complete space $S^1\times S^1$, the setsare closed. They cover that space: for a nonzero sequence in $c_0$, the supremum norm is attained at a coordinate; for $x=y$ every coordinate works. By the Baire category theorem, one $F_n$ contains a product of open arcs. Shrink the arcs to be disjoint. The real coordinate difference is then continuous and nonzero, so its sign is constant on the product. Thus the chord distance has the separated formIn particular its alternating sum on every four-point rectangle is zero. Parametrize the arcs by angles $s,t$ with $0<t-s<2\pi$. Their chord distance is $D(s,t)=2\sin ((t-s)/2)$, whose mixed derivative isThe zero rectangle identity would force this mixed derivative to vanish, a contradiction. Therefore no such isometry exists.

For a real vector $x\in {\ell }_1^n$, list its signed sums over all $2^n$ choices $\epsilon \in \{-1,1{\}}^n$, followed by infinitely many zero coordinates:This is linear, has finite support and lies in $c_0$. Choosing the signs of the components gives ${\max }_{\epsilon }|\epsilon \cdot x|={\sum }_j|x_j|$. Thus $\boxed{\Vert Tx{\Vert }_{\infty }=\Vert x{\Vert }_1}$, the required linear isometry.