Question on Reeb's theorem.

I am reading through the first couple of chapters of Milnor's Morse theory, and I've gotten to Reeb's sphere theorem (theorem 4.1),

If $$M$$ is a compact manifold and $$f$$ is a differentiable function on $$M$$ with only two critical points, both of which are nondegenerate, then $$M$$ is homeomorphic to a sphere.

Milnor states that for some small enough $$\varepsilon > 0$$, the sublevel sets $$f^{-1}[0,\varepsilon]$$ and $$f^{-1}[1-\varepsilon,1]$$ are closed $$n$$-cells, which follows by the Morse Lemma. But I don't quite follow this. I've read other related posts, but none of them quite flesh out why we need a "small enough" $$\varepsilon$$, or why exactly the Morse lemma implies that the aforementioned preimages are closed $$n$$-cells.

Now, I understand that the two critical points will have index $$0$$ and $$1$$, which correspond to the minimum and maximum, but why does applying the coordinate maps given to us my the Morse lemma tell us that we get $$n$$-cells, and not only containments? If anyone could clarify, that would be great.

Suppose that $$p$$ is a minimum of $$f$$ with $$f(p)=0$$. By Morse Lemma, there is a coordinate system $$(U,y^i)$$ of $$p$$ such that $$y^i(p) = 0$$ for all $$i=1,\dots,n$$ and $$f$$ has representation $$f = (y^1)^2 + \dots + (y^n)^2, \quad \text{on }U.$$ So, for sufficiently small $$\varepsilon>0$$ we have $$D \equiv f^{-1}[0,\varepsilon] \subseteq U$$. That is $$D = \{q \in U \, \mid \, f(y_q) = (y^1_q)^2+ \dots + (y^n_q)^2 \leq \varepsilon \},$$ which is clearly diffeomorphic to a closed $$n$$-disk in $$\mathbb{R}^n$$, via coordinate map $$y$$.