= Solution
We use <successive approximation for quotient operators>. Fix $y$ with $\|y\|<1$. The hypothesis gives $x_0$ with $\|x_0\|<1$ and $\|y-Tx_0\|<\delta$. Inductively, if
$$
r_n=y-T\sum_{k=0}^{n-1}x_k,\qquad\|r_n\|<\delta^n,
$$
apply the hypothesis to $r_n/\delta^n$, obtaining $x_n$ with $\|x_n\|<\delta^n$ and $\|r_n-Tx_n\|<\delta^{n+1}$. If a residual is zero, take all remaining terms zero.
The <series> $\sum x_n$ converges in the <Banach space> $X$ because $\sum\|x_n\|<\sum\delta^n=(1-\delta)^{-1}$. Write its sum as $x$. The <continuity> of $T$ and the residual estimate give
$$
Tx=y,\qquad\|x\|<(1-\delta)^{-1}.
$$
Scaling now yields
$$
\boxed{B_Y(0,1-\delta)\subseteq T(B_X(0,1)).}
$$
This ball inclusion proves that \b[$T$ is a <quotient operator>]. Completeness of $Y$ was unnecessary: the convergent <series> is constructed in $X$ and its <image> has limit $y$.
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