= Solution
Write $B$ for the <open unit ball>. Because $T$ is a <linear map>, the hypothesis scales to
$$
sB\subseteq\overline{T(sB)}
\qquad(s>0).
$$
Fix $y\in B$. Choose numbers $r$ and $\theta$ such that
$$
\lVert y\rVert<r<1,
\qquad 0<\theta<1-r;
$$
then $r/(1-\theta)<1$.
Set $z_0=y$. Inductively, if $\lVert z_{j-1}\rVert<r\theta^{j-1}$, the scaled density statement lets us choose
$$
x_j\in r\theta^{j-1}B
\quad\text{such that}\quad
\lVert z_{j-1}-Tx_j\rVert<r\theta^j,
$$
and we put $z_j=z_{j-1}-Tx_j$. The <geometric series> gives
$$
\sum_{j=1}^{\infty}\lVert x_j\rVert
<\sum_{j=1}^{\infty}r\theta^{j-1}
=\frac r{1-\theta}<1.
$$
Because $X$ is a <Banach space>, $x=\sum_{j\geq1}x_j$ exists and lies in $B$. Since $T$ is a <bounded linear operator>, it is continuous, while
$$
y-T\sum_{j=1}^Nx_j=z_N\longrightarrow0.
$$
Therefore $Tx=y$. As $y\in B$ was arbitrary,
$$
\boxed{T(B)\supseteq B.}
$$
This is the <dense open-unit-ball image criterion>.
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