= Solution
For $n\geq1$ put $C_n=\overline{T(n\overline B_X)}$, with <closure> taken in $Y$. Then
$$
T(X)=\bigcup_{n=1}^{\infty}T(n\overline B_X)\subseteq\bigcup_{n=1}^{\infty}C_n.
$$
If every $C_n$ has empty <interior>, then each $C_n$ is a <nowhere dense set>, because it is closed. Each $T(n\overline B_X)$ is then a <nowhere dense set> too. Consequently $T(X)$ is a <meagre set>, that is, a set of the first category in $Y$.
Otherwise some $C_n$ has nonempty <interior>. Scaling gives nonempty <interior> of $\overline{T(\overline B_X)}$. The <geometric correction for approximate surjectivity> proved in the preceding part gives an open <ball> about zero inside $T(\overline B_X)$. Since $T(X)$ is a <vector subspace>, it then equals $Y$: scale any $y\in Y$ into this <ball> and scale its preimage back. Hence $T(X)$ is a <closed set>.
Thus, in fact, the stronger <operator range category dichotomy> holds:
$$
\boxed{T(X)=Y\quad\text{or}\quad T(X)\text{ is meagre in }Y.}
$$
In particular $T(X)$ is closed or of the first category, as required. The only substantive theorem used here is the preceding <geometric correction for approximate surjectivity>: for a <bounded linear operator> from a <Banach space> to a <normed space>, nonempty <interior> of the closure of the image of the <unit ball> implies nonempty <interior> of the image itself. Its series proof was given there, so this proof does not require the <Baire category theorem> or completeness of $Y$.
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