= Solution
The <Baire category theorem> states that a nonempty complete <metric space> cannot be a countable union of closed <nowhere dense sets>; equivalently, every countable intersection of open dense sets is dense.
Let $E_n=\operatorname{span}\{e_1,\ldots,e_n\}$ in the space of <finitely supported sequences>. Each $E_n$ is a closed <linear subspace> for every <norm>, since a <finite-dimensional subspace is closed>. Hence
$$
F_n=T^{-1}(E_n)
$$
is a closed <linear subspace> of the <Banach space> $X$. Every <finitely supported sequence> lies in some $E_n$, so $X=\bigcup_{n\geq1}F_n$. By the <Baire category theorem>, some $F_N$ has nonempty <interior>.
A <linear subspace> with nonempty <interior> is the whole space: translate a ball in it to zero and scale any vector into that ball. Thus $F_N=X$, giving $T(X)\subseteq E_N$ and
$$
\boxed{\operatorname{rank}T\leq N<\infty.}
$$
The proof uses no particular choice of <norm> on the space of <finitely supported sequences>.
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