Schoen & Yau's proof of the positive-mass theorem: Why is the surface S homeomorphic to $\mathbb{R}^2$?

I'm currently reading through Schoen & Yau's 1979 proof of the positive-mass theorem and am trying to understand the following statement on p. 55 of the publication (page 11 of the proof / PDF):

Remark 2.1: The Cohn-Vossen inequality says that $$\int_S K \leq 2\pi \chi(S)$$, where [$$K$$ is the Gauss curvature and] $$\chi(S)$$ is the Euler characteristic of $$S$$. Combining this with (2.18) we see immediately that $$S$$ is homeomorphic to $$\mathbb{R}^2$$

Here, (2.18) refers to the inequality $$\int_S K > 0$$ derived just before.

How can I see that the surface $$S$$ is homeomorphic to $$\mathbb{R}^2$$? The immediate implication of (2.18) obviously is that $$\chi(S) > 0$$, so $$\chi(S) \geq 1$$. Apart from that, I know that $$S$$ is non-compact but I'm unsure about other properties$$^\dagger$$ of $$S$$ as my understanding of its construction is somewhat limited so far. Is it those properties that imply that the Betti numbers $$b_i$$ vanish for $$i \geq 1$$?

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$$^\dagger$$ For instance, I suspect that $$S$$ is boundaryless, connected & simply connected and also a closed subset of the ambient manifold $$N$$ but since I only have a rough idea of how $$S$$ is constructed as a limit of surfaces $$S_\sigma$$ (namely by representing the minimal surfaces $$S_\sigma$$ locally as graphs in the tangent space (using normal coordinates) and using Arzelà-Ascoli for finding the limit), I'm having trouble coming up with rigorous proofs of any properties of $$S$$ that could help prove the claim.

A noncompact surface has no homology groups above dimension 1. So $$\chi(S) = 1 - \dim H_1(S)$$. If $$\chi(S) = 1$$, then in fact $$H_1(S) = 0$$.
A noncompact surface with finitely generated homology groups is diffeomorphic to some $$S_{g,n}$$, the genus $$g$$ surface with $$n>0$$ points removed. This is proved using the similar classification of compact surfaces, and using a compact exhaustion of your manifold. I wrote something about the simplest case in the second part of the post here but it's not hard to extrapolate to the general case.
Now $$H_1(\Sigma_{g,n}) = \Bbb Z^{2g+n-1}$$ so long as $$n > 0$$. If this is zero (so that the Euler characteristic is 1), then necessarily $$g=0$$ and $$n=1$$. Puncturing the 2-sphere leaves you with $$\Bbb R^2$$, as desired.