= Solution
The real <Hahn-Banach separation theorem> says that if $C$ is a nonempty open <convex set> in a real <normed vector space> and $y\notin C$, there is a nonzero <continuous linear functional> $g$ such that $g(y)\geq\sup_{z\in C}g(z)$.
Put $\delta=1-1/M\in(0,1)$. The prescribed set is open because it is a union of translates of an open ball, and it is <convex> because both summands are <convex>. If $\|y\|<1$ and $y\notin C$, <Hahn-Banach separation> gives a nonzero $g\in Y^*$ with
$$
g(y)\geq\sup_{z\in C}g(z)=\|T^*g\|+\delta\|g\|\geq(M^{-1}+\delta)\|g\|=\|g\|.
$$
The displayed supremum is a sum of independent suprema; the symmetry of the unit balls identifies them with the respective <operator norms>. But $g(y)\leq\|g\|\|y\|<\|g\|$, a contradiction. Hence $y=Tx+r$ for some $\|x\|\leq1$ and $\|r\|<\delta$.
To obtain the strict bound on $x$ required in the approximation hypothesis, choose $t<1$ sufficiently close to $1$. Then $\|tx\|<1$ and
$$
\|T(tx)-y\|\leq\|r\|+(1-t)\|Tx\|<\delta.
$$
Thus the approximation condition holds with exactly $\delta=1-1/M$. When $X$ is a <Banach space>, the preceding <successive approximation for quotient operators> proves $T$ is a <quotient operator>. Combining this with the <Banach-space adjoint> estimate proves
$$
\boxed{T\text{ is a quotient operator}\iff T^*\text{ is an isomorphic embedding}.}
$$
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