Why are all open subsets not infinite in extent?

First, the definition that you have written is not correct. The sign '$\le$' should be '$<$'.

Now, to the answer. The set $U=\{z\in\Bbb C:|z|<1\}$ is open. To prove it, take any point $z_0\in U$. Since $|z_0|<1$, the number $r=\frac12(1-|z_0|)$ is positive, so we can define $$V=\{z\in\Bbb C: |z-z_0|<r\}$$

Then, for any $z\in V$ we have $$\begin{align} |z|&=|z-z_0+z_0|\le |z-z_0|+|z_0|\\ &<r+|z_0|=\frac12-\frac12|z_0|+|z_0|\\ &=\frac12+\frac12|z_0|<\frac12+\frac12=1 \end{align}$$

Therefore, $V\subset U$.

Geometrically (and informally) speaking, $V$ is an open disk centered at $z_0$ whose radius is small enough to be contained in $U$. Note that if we had written '$\le$' in the definition of $U$, then $U$ would have a border, and every disk centered on a point at the border would have a part inside $U$ and another part outside.