I’m going to comment only on what you wrote for (i):

Assume for all $x_n \in X$ with $\| x_n \|_X = 1$ that

$$x_n^\ast (x_n) \lt \frac{\| x_n^\ast \|}{2} \iff 2 x_n^\ast (x_n) \lt \sup_{x_n; \| x_n \|_X = 1 } \{| x_n^\ast(x_n)|\}$$

Claim: Then $\exists r \in \mathbb{R}: x_n^\ast(x_n) \lt r \lt \sup_{\dots}\{\dots \}$:

(i) $x_n^\ast (x_n) \neq 0$: because $\| x_n \| = 1$ and $x_n^\ast$ linear

(ii) if $x_n^\ast (x_n) \lt 0$ then $2 x_n^\ast (x_n) \lt x_n^\ast (x_n) \lt x_n^\ast(- x_n) \lt 2 x_n^\ast(-x_n) \lt \sup_{\dots} \{ \dots \}$

(iii) $x_n^\ast(x_n) \gt 0$ then $\forall x_n: x_n^\ast (x_n) \lt 2 x_n^\ast (x_n) \lt \sup$

$\implies $ the $\sup$ is not the l.u.b., contradiction, so the claim $x_n^\ast (x_n) \geq \frac{\| x_n^\ast \|}{2}$ is true.

The first problem, and it’s a major one, is that right off the bat you’re using $n$ for two completely different things. On the one hand it’s apparently supposed to be the index of a particular, fixed member of some countable norm-dense subset of $X^*$, though you never actually said that. On the other hand it’s a dummy index picking out members of $X$ of norm $1$; this would be a bad idea even if you weren’t already using $n$ for something else, since there’s no reason to suppose that there are only countably many elements of $X$ of norm $1$. You should have begun something like this:

Let $D=\{x_n^*:n\in\mathbb{N}\}$ be a countable norm-dense subset of $X^*$. Fix $x_n^*\in D$, and assume for each $x\in X$ with $\|x\|=1$ that $$x_n^*(x)<\frac{\|x_n^*\|}2.$$

Note that I stopped *before* your $\iff$ symbol: that’s because what follows it is *not* part of your assumption, but rather an inference *from* your assumption, so it does not belong in the same clause with *assume that*. Once you’ve clearly stated your assumption, *then* you can go on and draw conclusions:

This implies that $$2x_n^*(x)<\sup_{\|y\;\|=1}\|x_n^*(y)\|$$ for each $x\in X$ with $\|x\|=1$.

Note that I had to use a different letter for the dummy variable ($y$) in the supremum from the one used for the specific $x$ of norm $1$ in the surrounding statement: they refer to different objects.

The line that begins *Claim* makes no sense even after the ellipsis at the end is properly filled in. (Note, by the way, that this is something that you should have done yourself, so that the reader needn’t guess; with cut-and-paste it’s completely trivial.) You still have $n$ meaning two different things, and it’s not clear whether you’re talking talking about a specific $x$ of norm $1$ or all of them together. I suspect that you meant this:

**Claim:** There is an $r\in\mathbb{R}$ such that $$x_n^*(x)<r<\sup_{\|y\;\|=1}|x_n^*(y)|$$ for each $x\in X$ of norm $1$.

Presumably what follows is supposed to be a proof of the claim. **Say so.**

**Proof of Claim:** Fix $x\in X$ of norm $1$. Since $\|x_n^*\|$ is linear, $x_n^*(x)\ne 0$.

This doesn’t appear to follow. In fact, there seems to be nothing in your argument to here to preclude the possibility that $x_n^*$ is the zero functional.

If $x_n^*(x)>0$, then $$x_n^*(x)<2x_n^*(x)<\sup_{\|y\;\|=1}|x_n^*(y)|\;,$$ and if $x_n^*(x)<0$, then $$2x_n^*(x)<x_n^*(x)<x_n^*(-x)<2x_n^*(-x)<\sup_{\|y\;\|=1}|x_n^*(y)|\;.$$

This does not in fact prove the claim; you’ve merely shown that for each $x\in X$ of norm $1$, $$x_n^*(x)<\|x_n^*\|=\sup_{\|y\;\|=1}|x_n^*(y)|\;,$$ i.e., that $x_n^*$ does not attain its norm on $\{y\in X:\|y\|=1\}$.

Had you made the effort to write it intelligibly, thinking about what you were actually saying, you might have noticed that much of it makes no sense and that it does not in fact do what you wanted it to do. At the very least you would have made it possible for others to spot the problems and help with them *much* more easily.

**Remember: A proof is just a particular kind of expository prose. It should consist of paragraphs of sentences. Yes, it will often contain special symbols, but DON’T use symbols just for the sake of using them. The object is to convince the reader that something is true, and you can’t do that if you can’t make yourself understood.**

~~typo~~`. It would have been much better if you had appended the new bit to the question indicating clearly that it is new. $${}$$ Concerning the math: there's still the notational problem that the $n$ has two completely different meanings in the beginning of the argument. Replace $x_n$ by $x$ there. I would start: If $x_{n}^\ast = 0$ then we can take $x_n$ to be any vector of norm one... – t.b. Nov 16 '11 at 13:26