Definition: A subset $Z$ in the normed linear space $X$ is said to be fundamental if $span(Z)$ is dense in $X$.
This is the statement of Theorem 1.6.3 in Chaney's Analysis for Applied Mathematics.
A subset in a normed linear space is fundamental if and only if its annihilator is $\{0\}$.
Now, I understand the first part of the given proof which uses (a corollary of) the Hahn-Banach Theorem. Next, denoting the closure of the $span(Z)$ by $Y$, this is the second direction of the proof as given in the book:
If $Y=X$, then any element of $Z^\perp$ annihilates the span of $Z$ as well as $Y$ and $X$. Thus it must be the zero functional; i.e., $Z^\perp = \{ 0 \}$.
It is easy to show that the annihilator of $Z$, $Z^\perp :=\{\phi \in X^*, \phi(z)=0, \forall z \in Z \}$ is equal to $span(Z)^\perp$. But I do not see how the part "...as well as $Y$ and $X$" is concluded. I'd appreciate any explanation. Thanks.
Edit: So may I argue as below?
$\forall y \in Y = \overline{span(Z)}, \exists$ a sequence $[y_n]$ in $span(Z)$ such that $y_n \rightarrow y $. Now, $\forall \phi \in Z^\perp, \forall y_n \in span(Z), \phi(y_n)=0$. Thus, because $\phi$ is continuous, we get $\phi(y)=0$, making every such $\phi$ a member of $Y^\perp$, as well.