Analogue of $S^{\bot \bot} = \overline{S}$ for Banach spaces which aren't Hilbert spaces Let $X$ be a Banach space over the complex field. Let $X^*$ denote its topological dual. If $S$ is a subspace of $X$, write
$$\mathrm{ann}_L(S)= \{ \varphi \in X^* : \varphi S = 0\}$$
and note the result is a weak-star closed subspace of $X^*$. Similarly, if $S$ is a subspace of $X^*$, write
$$\mathrm{ann}_R (S) = \{ x \in X : S x = 0 \}$$
and note the result is a weakly closed subspace of $X$. It is obvious that $\mathrm{ann}_L$ and $\mathrm{ann}_R$ are order-reversing. Also, if $S, T$ are subspaces of $X, X^*$ respectively, we have
$$ S \subset \mathrm{ann}_R(T) \Leftrightarrow T \subset \mathrm{ann}_L(S) \Leftrightarrow TS = 0$$
so that $\mathrm{ann}_L$ and $\mathrm{ann}_R$ set up an (antitone) Galois connection between the posets of subspaces of $X$ and $X^*$. Various things then follow by abstract nonsense. For instance, $\mathrm{ann}_R \circ \mathrm{ann}_L$ and $\mathrm{ann}_L \circ \mathrm{ann}_R$ are abstract closure operators and the associated "closed subspaces" of $X$ and $X^*$ are put into order-reversing bijection by $\mathrm{ann}_L$ and $\mathrm{ann}_R$. In light of the fact that the range of $\mathrm{ann}_L$ consists of weak-star closed subspaces and the range of $\mathrm{ann}_R$ consists of weakly closed subspaces, it is natural to wonder whether these closure operators are, in fact, equal to the weak and weak-star closure operators. In essence, I am asking the following.

Question 1 (answered): Let $S$ be a subspace of $X$ and let $x \in X$. If $\varphi S = 0$ implies $\varphi(x) =0$ for all $\varphi \in X^*$, does it follow that $x$ is in the weak closure of $S$?
Question 2: Let $S$ be a subspace of $X^*$ and let $\varphi \in X^*$. If $S x= 0$ implies $\varphi(x) =0$ for all $x \in X$, does it follow that $\varphi$ is in the weak-star closure of $S$?

Thank you in advance for any answers or clarification on surrounding issues.
Added: I've managed to answer Question 1 affirmatively. First recall that the weak closure of $S$ is the same as the norm closure of $S$. More generally, the weak closure and norm closure coincide for convex subsets of $X$ (Conway, A Course in Functional Analysis, Theorem V.1.4). The Hahn-Banach Theorem implies the following statement: If $S \subset X$ is a subspace and $x \in X$ is a positive distance $d$ away from $S$, then there exists a functional $\varphi \in X^*$ with $\|\varphi\| = 1$ and $\varphi(x) = d$ and $\varphi S = 0$. It is easy to answer Question 1 using these two facts.
 A: The key for question 2 is to observe that the weak-star continuous functionals on $X^\ast$ are precisely the evaluation functionals at points $x \in X$. From there the answer proceeds similarly to what you did in your answer for question 1:
If $\varphi$ is not in the weak-star closure $\overline{S}^{w\ast}$ of $S$ then Hahn-Banach applied to $X^\ast$ with the weak-star topology ensures that we can separate $\varphi$ from $\overline{S}^{w\ast}$, so there must be $x \in X$ such that $\varphi(x) \gt s(x)$ for all $s \in \overline{S}^{w\ast}$. Since $\overline{S}^{w\ast}$ is a linear subspace, we must have that $x \in \mathop{{\rm ann}_R} S$ and hence $\varphi \notin \mathop{{\rm ann}_L} \mathop{{\rm ann}_R} S$. This shows $\mathop{{\rm ann}_L} \mathop{{\rm ann}_R} S \subseteq \overline{S}^{w\ast}$ while $S \subseteq \mathop{{\rm ann}_L} \mathop{{\rm ann}_R}S$ and weak-star closedness of the latter shows the other inclusion.
A good reference for this and related results is Rudin's Functional Analysis, chapter IV on duality of Banach spaces.
