Non-compact Riemann surface has only one end

I'm trying to understand proof for the following result:

Let $X$ be a connected, simply connected, non-compact surface. Then, for any compact set $K\subset X$, the complement $X\setminus K$ has exactly one connected component whose closure is not compact.

As a side note I think we can assume that 'surface' means 'Riemann surface' in every occurrence.

So this can be found in Simon Donaldson's Riemann Surfaces on page 141: https://books.google.fi/books?id=3SwSDAAAQBAJ&printsec=frontcover&hl=fi#v=onepage&q&f=false

In the fourth paragraph, after construction of compactly supported 2-forms $\sigma_i$ we can find compactly supported 1-form $\alpha$ such that $d\alpha = \sigma_1 - \sigma_0$. The text refers to Proposition 12 but the correct result is Proposition 14 on page 70:

If $S$ is a connected, oriented, smooth surface, then the map $\int_S$ is an isomorphism from $H^{2}_{c} (S)$ to $\mathbb{R}$.

Now we choose a neighbourhood $N$ and the problems start. If $N$ is contained in $K$ then either $N$ is a strict subset of $\text{supp}(\alpha)$ as $\alpha$ must be non constant on the support of $\sigma_i$ or both of the sets $N$ and $\text{supp}(\alpha)$ have points not belonging to other set. Now I don't see why we can find a loop $\gamma$ in $N$ so that $\int_\gamma \alpha=1$, because all that I know is that $\alpha$ is non-zero on the supports of $\sigma_i$. Okay, assume that 'by some magic' we can find said $\gamma$. Towards the end of the proof, the fact that $\gamma$ does not meet the support of $\beta_i$ is used. By construction of $\beta_i$ the supports of $\beta_i$ and $\sigma_i$ intersect. So what I had in mind to fix this does not work. I thought that if we make a union of the supports of $\sigma_i$ and $K$ and make the union into a connected set with the help of paths and covering of path by open sets that have compact closures. That would make the choice of $\gamma$ more plausible, but the arguments at the end of the proof make this enlargement of $K$ unusable.

So what am I missing or is there really something wrong with the proof?

$\renewcommand\supp{\operatorname{supp}}$For the first part about $N$, the point is that we can choose any neighbourhood $N$ of $\supp\theta$ contained in $K$. (I think that for this to exist, we need $f = 0$ on a neighborhood of $\overline{U_0}$, not just in $U_0$, and likewise for $f=1$ and $U_1$.) Then, since $\supp (d\alpha) \subseteq U_0 \cup U_1 \subseteq S \smallsetminus N$, the restriction $d\alpha|_N$ of $d\alpha$ to $N$ is zero. This is what he means by $\alpha$ being a closed form on $N$. (I wonder if this is the point of confusion for you.) The support of $\alpha$ itself is irrelevant.
Now he shows $\alpha|_N$ is not exact, so it represents a nonzero cohomology class in $H^1(N)$. This implies the existence of a loop $\gamma$ in $N$ with $\int_\gamma \alpha \ne 0$, for otherwise if $\int_\gamma \alpha = 0$ for any loop $\gamma$ in $N$, we could construct a $0$-form $u$ on $N$ with $du = \alpha|_N$ by integrating $\alpha$. Whether the integral is $1$ or some other nonzero value doesn't matter in the rest of the proof.
• So now $N$ cannot be simply connected and by simple argument neither can $K$ be simply connected. We wanted to choose any compact set, so this is our desired contradiction. – VadaVad Oct 4 '16 at 14:11