What does "that is, on maximal ideals of $k[x]$" mean? Just before that remark it was said that we should work in the Zariski topology on $A^1(k)$, so the remark following is confusing. What exactly is being asked?
Assuming that the question is "show that the open subsets of $A^1(k)$ are the complements of finite sets": an open subset of $A^1(k)$ is of the form $A^1(k)\setminus Z(I)$ for an ideal $I\subset k[x]$. It is indeed the complement of the finite set $Z(I)$: since $k$ (hence $k[x]$) is Noetherian, $I$ is finitely generated, and every generator has only finitely many roots. Is that reasoning correct?
Conversely, suppose $S$ is a finite subset of $A^1(k)$. I need to show that $A^1(k)\setminus S$ is open. So I need to show that $S$ is an algebraic set, right? Shall I just consider the ideal generated by a polynomial that has as roots the elements of $S$ (e.g. the corresponding Lagrange polynomial)? This should prove that $S$ is the zero set of that ideal, so $A^1(k)\setminus S$ is open. But what for do we need that $k$ is algebraically closed?
For the last question, should I prove it for all $n$, or is it suffices to prove it for $n=2$? For $n=2$, this answer suggests to look at $x=y$, but I don't see why this is a counterexample: the zero set of $x-y$ is closed in the Zariski topology, and it's also closed say in $\mathbb R^2$ (the product topology coincides with the Euclidean topology, and the line contains all its limit points, hence is closed).