I have been looking at the definition of an open set, for a metric space. I have come across the following definition, a few times:
An open set $U$ of the metric space $(X,d)$ is a set given that $\forall x \in U$ there is at least one $\epsilon>0$ such that $\{y|y\in X, d(x,y)\le\epsilon\}\subseteq U$
To me this means is no final point at the edge of the set, since there always has to be one more for the condition above to hold. Hence the set cannot have an 'edge' and thus must be infinite in extent. Why is my reasoning wrong.
(I am looking at this from the complex plane, where the definition says something along the lines of 'for an open set of the complex plane' which to me seems like we can't have an open set of the complex plane, unless it is the whole complex plane)
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.