This is theorem 15.1 on Hamilton's "Three manifolds with positive Ricci curvature".
I don't understand why he says "thus we can reach every point of X by a geodesic of length at most...". I mean, he proved the estimate $R\geq (1-\eta)R_{max}$ for every point in the geodesic ball centred at $x$ with length $s=\frac{1}{\eta \sqrt{R_{max}}}.$ Then he used Myers' theorem to show that you can actually find a conjugate point within the distance $s.$ How this leads him to conclude that this geodesic ball is actually all of $X,$ i.e every point of $X$ can be reached by a geodesic of length at most $s$?
For any point $y \in X,$ let $\gamma : [0,1] \to X$ be a length-minimizing segment joining $x$ to $y,$ so that $L(\gamma) = d(x,y).$ If $\gamma$ had length greater than $s,$ then it would have a conjugate point in its interior, which contradicts the fact that it is length-minimizing; so we have $d(x,y) \le s.$ Since $y$ was arbitrary, we conclude $B(x,s) = X.$