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The theorem is as follows

Theorem 3.26. Let $E$ be a Banach space such that $E^*$ is separable. Then $E$ is separable.

Proof. Let $(f_n)$ be countable and dense in $E^*$. Since \begin{align*} \|f_n \| = \sup_{x \in E, \|x\| \leq 1} \langle f_n, x \rangle, \end{align*} we can find some $x_n \in E$ such that \begin{align*} \|x_n \| = 1 \text{ and } \langle f_n, x_n \rangle \geq \frac{1}{2} \|f_n\| \end{align*}

I am immediately stuck at this point in the proof. Why can we always find such an $x_n$?

To reason about this, I supposed by contradiction that for any $x \in E$ such that $\|x\| = 1$, we have $\langle f_n, x \rangle < \frac{1}{2} \|f_n \|$. Taking the sup over such $x$, we have $\sup_{\|x\| = 1} \langle f_n,x\rangle \leq \frac{1}{2} \|f_n \|$. By definition of sup, we can always find a $y$ s.t. $\| y \| \leq 1$ and $\langle f_n, y\rangle > \|f_n\| - \epsilon$ for arbitrary $\epsilon > 0$. Then $\frac{1}{2}\|f_n\| < \frac{1}{2} \langle f_n, y\rangle + \frac{\epsilon}{2} \Rightarrow \sup_{\|x\| = 1} \langle f_n,x\rangle < \frac{1}{2} \langle f_n, y\rangle + \frac{\epsilon}{2}$. If $\|y\| = 1$, then we have a contradiction since $\epsilon$ is arbitrary. I don't know how to derive a contradiction if $\| y \| < 1$.

Am I going on the right track? There must be a better way to see this.

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    Don't forget the alternate definition $||f_n|| = \sup_{||x||=1} \langle f_n,x \rangle$. In the sentence: "Taking the sup over such $x$ ...", the statement next directly translates to $||f_n|| \leq \frac 12 ||f_n||$, which is only possible if $||f_n|| = 0$, but this isn't true, that is the contradiction. – 2017-01-12
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    It follows from the definition of a supremum. – 2017-01-12

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Note that the implicit assumption here is that the inner product $\langle\cdot,\cdot\rangle$ is real-valued (otherwise sup does not make sense). Definition of sup tells us given $f\in E^*$ we can find $x\in E$ such that $$ \|x\|\le1\qquad\text{and}\qquad\langle f,x\rangle>\frac{1}{2}\|f\| $$ If $\|x\|=1$ we're done otherwise define $x^*=x/\|x\|$. Now $\|x^*\|=1$ and according to one of the defining properties of a real-valued inner product (namely linearity in second argument) we have $$\langle f,x^*\rangle=\frac{1}{\|x\|}\langle f,x\rangle>\frac{1}{2\|x\|}\|f\|>\frac{1}{2}\|f\|\quad(\because\|x\|<1)$$