So, this is actually correct. For any subset $S$ of an inner product space $V$, we do have
$$\operatorname{\overline{span}} S = (S^\perp)^\perp.$$
In terms of where you went wrong with your example, I think you have an issue with your understanding about the perpendicular complement. We define
$$S^\perp = \{v \in V : \langle v, w \rangle = 0\}.$$
In particular, if $S$ is the $x$-axis, i.e. $S = \{(x, 0) : x \in \Bbb{R}\}$, then $(a, b) \in S^\perp$ if and only if
$$(a, b) \cdot (x, 0) = 0$$
for all $x \in \Bbb{R}$. In particular, if we consider $x = 1$, we require
$$0 = (a, b) \cdot (1, 0) = a,$$
i.e. $(a, b)$ takes the form $(0, b)$. That is, $(a, b)$ lies on the $y$-axis.
In particular, note that this is not the set of lines parallel to the $y$-axis, it is just the $y$-axis itself. Remember, a $0$ inner product means that the vectors are perpendicular from the origin.
Similarly, of course, $(S^\perp)^\perp$ is the $x$-axis. It's just a collection of vectors which are perpendicular to the $y$-axis, starting at the origin.
