I try to prove the following:
Let $\hat{\mathbb C}$ the Riemann sphere with the chordal metric $\hat d$. Then $U \subseteq \hat{\mathbb C}$ is an open neighbourhood of $\infty$ if and only if $U = \hat{\mathbb C} \setminus K$ for some compact set $K \subseteq \mathbb C$.
But I am a little suspicious about my solution for some reason:
"$\Rightarrow$": Let $U$ be an open neighbourhood of $\infty$. Then $\infty \in U$ and thus $K := \hat{\mathbb C} \setminus U \subseteq \mathbb C$ is closed. Since $\hat{\mathbb C}$ is a compact metric space it follows that $K$ is compact with respect to $\hat d$. Since the usual euclidean metric $d$ and $\hat d$ are topologically equivalent it follows that $K$ is also compact to with respect to $d$.
"$\Leftarrow$": Let $U = \hat{\mathbb C} \setminus K$ for some compact set $K \subseteq \mathbb C$. Then obviously $\infty \in U$. Further since $d$ and $\hat d$ are topologically equivalent it follows that $K$ is compact with respect to $\hat d$. Since $\hat{\mathbb C}$ is Hausdorff $K$ is closed with respect to $\hat d$ and $U =\hat{\mathbb C} \setminus K$ thus open.
Is this the right way to solve it or are there flaws in my arguments?
This is quite correct. The metric is not really relevant as you can see, hence this reformulation in terms of compact complements. Note that even $\infty$ is not special anymore, any point on the Riemann sphere has open neighbourhoods that have compact complement (in the Riemann sphere), just because we're in a compact Hausdorff space. The only thing that marks out $\infty$ is that the compact complement "lives" entirely on $\mathbb{C}$. Look at wikipedia for the general construction for locally compact Hausdorff spaces.