Can Hahn-Banach separation theorem in $X^{\ast}$ be generalized in $w^{\ast}$ topology

When reading a proof of this theorem (which states every uniformly convex Banach space is reflexive), I came across the following claim:

In a infinite-dimensional Banach space $$X$$, if $$C$$ is a weak-$$\ast$$ compact subset of $$X^{\ast\ast}$$ and $$z$$ is a point disjoint from $$C$$, then we can find $$g \in X^{\ast}$$ such that $$g$$ strictly separates them.

My questions are:

1. By Hahn-Banach Separation Theorem we could find $$\phi \in X^{\ast\ast\ast}$$ and $$\alpha \in \mathbb{R}$$ such that $$\sup_{x \in C}\operatorname{Re}\phi(c) < \alpha < \operatorname{Re}\phi(z)$$. Can we find such $$g \in X^{\ast}$$ that can replace $$\phi$$? I suppose the author wants to cite Goldstine's Theorem but in this case the set $$\{\psi \in X^{\ast\ast\ast}\,\vert\,\vert\psi(x) - \phi(x)\vert < \epsilon\,\forall\,x \in C\} = \bigcap_{x \in C}\{\psi \in X^{\ast\ast\ast}\,\vert\,\vert\psi(x) - \phi(x)\vert < \epsilon\}$$ may not be open and how can we find $$g \in X^{\ast}$$ that is close enough to $$\phi$$ on every point in $$C$$?
2. Given a infinite-dimensional Banach space $$X$$, in $$X^{\ast}$$ equipped with the weak-$$\ast$$ topology are there any sufficient conditions to finding an injective bounded linear functional (an example will be greatly appreciated)? How about in a general topological vector space? (I suppose a nice mapping between the basis in an neighborhood of $$\overset{\rightarrow}{0}$$ and a bounded set in $$\mathbb{R}$$ always exists but hope to a see some details).

1. Yes. In general, let $$Y$$ be a normed space and endow $$Y^*$$ with the weak-star topology, which is a locally convex topology. If $$C\subset Y^*$$ is a convex, weak-* closed set and $$z\in Y^*\setminus C$$, then by the Hahn-Banach theorem we can find a real number $$t$$ and a functional $$\phi:Y^*\to\mathbb{C}$$ that is continuous $$\textbf{with respect to the weak-star topology}$$ such that $$\text{Re}(\phi(c)) for all $$c\in C$$. I claim that the functionals $$\phi:Y^*\to\mathbb{C}$$ that are continuous with respect to the weak-star topology are precisely the evaluation functionals of points of $$Y$$, i.e. precisely the functionals of the form $$\text{ev}_y:Y^*\to\mathbb{C}$$, $$f\mapsto\text{ev}_y(f):=f(y)$$.
To see this, let $$\phi:Y^*\to\mathbb{C}$$ be continuous for the weak-star topology. Since the weak-star topology is generated by the seminorms $$\{p_y\}_{y\in Y}$$ where $$p_y(f)=|f(y)|$$, we can find $$M>0$$ and $$y_1,\dots,y_n\in Y$$ such that $$|\phi(f)|\leq M\cdot\max_{1\leq i\leq n}|p_{y_i}(f)|$$ for all $$f\in Y^*$$, i.e. $$|\phi(f)|\leq M\max_{1\leq i\leq n}|f(y_i)|$$ for all $$f\in Y^*$$. But this shows that $$\bigcap_{i=1}^n\ker(\text{ev}_{y_i})\subset\ker(\phi)$$. By elementary linear algebra this shows that $$\phi$$ is a linear combination of $$\text{ev}_{y_i}$$, thus of the form $$\text{ev}_y$$, where $$y$$ is a linear combination of the $$y_i$$.
Therefore, we can restate the application of Hahn-Banach as "we find a real number $$t$$ and a point $$y\in Y$$ such that $$\text{Re}(c(y)) for all $$c\in C$$.
1. Let $$X$$ be a vector space in general. If we have an injective linear map $$T:X\to\mathbb{C}$$, then $$X$$ embeds as a vector space in $$\mathbb{C}$$, so $$\dim(X)\leq1$$. The condition you are looking for is this: $$X$$ admits an injective linear functional if and only if $$X$$ is trivial or one-dimensional.