= 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>. Thus
$$
\boxed{\bigl|\{U\subseteq\mathbb C:U\text{ open}\}\bigr|=\mathfrak c.}
$$
The countable-basis restriction is decisive: there are $2^{\mathfrak c}$ arbitrary <subsets> of the plane, but only continuum many open ones.
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