= Solution
A <quasi-isometry> $f:X\to Y$ has constants $\lambda\ge1$, $\epsilon\ge0$ such that
$$
\lambda^{-1}d_X(x,x')-\epsilon\le d_Y(f(x),f(x'))\le\lambda d_X(x,x')+\epsilon
$$
for all $x,x'$, and every $y\in Y$ is within $\epsilon$ of some point of $f(X)$. One may use a separate coarse-surjectivity constant and enlarge $\epsilon$ to include it. No continuity is required.
Choose $f'(y)\in X$ with $d_Y(f(f'(y)),y)\le\epsilon$. The triangle inequality gives
$$
d_Y(y,y')-2\epsilon\le d_Y(f(f'(y)),f(f'(y')))\le d_Y(y,y')+2\epsilon.
$$
Combining this with the distortion inequalities for $f$ yields
$$
\lambda^{-1}d_Y(y,y')-\frac{3\epsilon}{\lambda}\le d_X(f'(y),f'(y'))\le\lambda d_Y(y,y')+3\lambda\epsilon.
$$
Also $d_Y(f(f'(f(x))),f(x))\le\epsilon$, so the lower distortion inequality gives $d_X(f'(f(x)),x)\le2\lambda\epsilon$. Hence every point of $X$ is within $2\lambda\epsilon$ of the image of $f'$. Taking the larger common error constant shows that \b[$f'$ is a $(\lambda,3\lambda\epsilon)$ <quasi-isometry>]. It is a <quasi-inverse of a quasi-isometry>; both composites are uniformly close to their respective identities.
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