It is not true that a finer cover necessarily contains more open sets than a coarser one. Consider $\Bbb Z$ with the discrete topology: $\mathscr{U}=\{\Bbb Z\}$ is an open cover of $\Bbb Z$ with one member, $\mathscr{V}=\big\{\{n\}:n\in\Bbb Z\big\}$ is an open cover of $\Bbb Z$ with (countably) infinitely many members, yet $\mathscr{V}$ is a refinement of $\mathscr{U}$ and therefore finer than $\mathscr{U}$. In the other direction, $\mathscr{W}=\wp(\Bbb Z)\setminus\{\varnothing\}$ is an uncountable open cover of $\Bbb Z$, and $\mathscr{U}$ and $\mathscr{V}$ are both refinements of $\mathscr{W}$, so a finer cover can have fewer members. This notion of finer and coarser really has nothing at all to do with the cardinalities of the covers.
In both contexts the intuition behind the terms finer and coarser is that a finer topology or cover chops the space up more finely. If $\tau_1\subseteq\tau_2$, where $\tau_1$ and $\tau_2$ are topologies on some set $X$, then $\tau_2$ chops up $X$ into all of the (open) pieces that $\tau_1$ does and possibly more besides. In this case that actually does imply that $|\tau_2|\ge|\tau_1|$, but that cardinality inequality is a byproduct, not the basis for the terminology. Similarly, if $\mathscr{V}$ is a refinement of $\mathscr{U}$, then any member of $\mathscr{V}$ containing some $x\in X$ is at least as small as some member of $\mathscr{U}$: if $x\in V\in\mathscr{V}$, then there is a $U\in\mathscr{U}$ such that $x\in V\subseteq U$.
There are differences, of course: a finer topology always has all of the open sets of the coarser topology, and possibly more besides, while a refinement need not contain any member of the original cover. But in both cases the ‘finer’ object does in some meaningful sense divide up the underlying set in a more fine-grained fashion.
Let $Z\subset\mathbb{A}^n=X$ a closed subset; by definition
$$
\exists f_1,\dots,f_r\in\mathbb{K}[x_1,\dots,x_n]=R\mid Z=V(f_1,\dots,f_r)=\bigcap_{i=1}^rV(f_r),
$$
withous loss of the generality we can assume $r=1$ and we put $f_1=f$ so that $Z=V(f)$; because $R$ is an U.F.D., then
\begin{gather*}
\exists g_1,\dots,g_s\in R\mid f=g_1\cdot\dots\cdot g_s,\,g_j\,\text{is a prime polynomial}\\
Z=V(g_1\cdot\dots\cdot g_s)=\bigcup_{j=1}^sV(g_j),
\end{gather*}
without loss of generality we can assume $s=1$, in other words $f$ is a prime polynomial and $Z$ is an irreducible closed subset of $X$.
Let $\{U_i\}_{i\in I}$ an open covering of $Z$, for exact:
$$
Z\subseteq\bigcup_{i\in I}U_i,
$$
by previous reasoning, we can assume (without loss of the generality) that $U_i$'s are irreducible; by definition:
\begin{gather*}
\forall i\in I,\exists f_i\in R\mid U_i=D(f_i)=X\setminus V(f_i),\\
\bigcup_{i\in I}U_i=\dots=X\setminus\bigcap_{i\in I}V(f_i)=X\setminus W;
\end{gather*}
we know that $W$ is a closed subset of $X$, let $I(W)$ be the associated ideal of $W$, by Hilbert's Base theorem, it is finitely generated; by this statement
\begin{gather*}
\exists I_F\subseteq I\,\text{finite,}\,\{f_i\in R\}_{i\in I_F}\mid I(W)=(f_i\mid i\in I_F)\Rightarrow\\
\Rightarrow Z\subseteq\bigcup_{i\in I}U_i=X\setminus V(I(W))=X\setminus\bigcap_{i\in I_F}V(f_i)=\bigcup_{i\in I_F}U_i
\end{gather*}
and the claims follows. (Q.E.D.) $\Box$
Remark: I had use only the hypothesys that $X$ is an affine space over a field, independently from its characteristic and other algebraic properties!
Best Answer
It's a good first step to understand the version with $\mathbb{C}$ in place of $\mathbb{C}^2$. Here we're looking at single-variable polynomials over $\mathbb{C}$, and these are relatively simple. In particular, we have a good understanding of $\{u: f(u)=0\}$ for such an $f$:
This means that any closed in the usual sense subset of $\mathbb{C}$ not satisfying this same "size condition" cannot be Zariski closed. For example:
OK, now how can we lift this to $\mathbb{C}^2$?
Well, there are various ways to do this, but one I quite like is to consider sections. Suppose $f(u,v)$ is a polynomial over $\mathbb{C}$ in two variables. Fix some $z\in\mathbb{C}$; we then get a single-variable polynomial $$g(u)=f(u,z).$$
What can we say about $\{u: g(u)=0\}$ (thinking about the previous section of this answer?
How does that give an example of a subset $A\subseteq\mathbb{C}^2$ which is closed in the usual topology but not in the Zariski topology?