 Solution

The complex plane has a countable basis of disks with positive rational radii and centres with rational real and imaginary coordinates. Each open set is the union of the basic disks it contains, so it is determined by a subset of this countable basis. Therefore the number of open sets is at most $2^{{\aleph }_0}=\mathfrak{c}$.

For the reverse inequality, the distinct disks $\{z:|z|<r\}$ for $r>0$ give $\mathfrak{c}$ different open sets. ThusThe countable-basis restriction is decisive: there are $2^{\mathfrak{c}}$ arbitrary subsets of the plane, but only continuum many open ones.