 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 givesWrite $\delta =r/2$. By scaling,Fix $y$ with $\Vert y\Vert <\delta /2$. Starting with $r_0=y$, recursively choose $x_j\in X$ such thatAt 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 ${\sum }_j{x}_j$ converges to some $x$ because $X$ is a Banach space, and the triangle inequality gives $\Vert x\Vert \le {\sum }_j{2}^{-j}=1$. Continuity of $T$ gives $Tx={\lim }_N{\sum }_{j=1}^{N}T{x}_j=y$, since $r_N\to 0$. ThusScaling by $1/2$ now gives the slightly smaller open ballThis 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.