= Solution
Use the <closed unit ball> $\overline B_X$ first, and put $C=\overline{T(\overline B_X)}$. This is a <convex set> symmetric about zero. If $B_Y(y_0,r)\subseteq C$, then $-y_0\in C$, so taking midpoints gives
$$
B_Y(0,r/2)\subseteq C.
$$
Write $\delta=r/2$. By scaling,
$$
B_Y(0,2^{-j}\delta)\subseteq\overline{T(2^{-j}\overline B_X)}\qquad(j\geq1).
$$
Fix $y$ with $\|y\|<\delta/2$. Starting with $r_0=y$, recursively choose $x_j\in X$ such that
$$
\|x_j\|\leq2^{-j},\qquad
r_j=r_{j-1}-Tx_j,\qquad
\|r_j\|<2^{-j-1}\delta.
$$
At each step the previous residual lies in the open <ball> on the left of the scaled inclusion, so the definition of <closure> permits this approximation. The <series> $\sum_jx_j$ converges to some $x$ because $X$ is a <Banach space>, and the <triangle inequality> gives $\|x\|\leq\sum_j2^{-j}=1$. Continuity of $T$ gives $Tx=\lim_N\sum_{j=1}^NTx_j=y$, since $r_N\to0$. Thus
$$
B_Y(0,\delta/2)\subseteq T(\overline B_X).
$$
Scaling by $1/2$ now gives the slightly smaller open <ball>
$$
\boxed{B_Y(0,\delta/4)\subseteq T(B_X(0,1)).}
$$
This proves the assertion whether the original <unit ball> is taken open or closed. Only completeness of $X$ was used; the <normed space> $Y$ need not be complete. The argument is <geometric correction for approximate surjectivity>.
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