Prove that $ \|x\|=\sup\{|f(x)|:f\in X^*, \,\|f\|=1\},$ where $x\in X$ and $X^*$ denotes the dual space of $X$. Let $x$ be an element of a normed linear space $X$ and let $X^*$ denote the dual space of $X$. Prove that 
\begin{align} \|x\|=\sup\{|f(x)|:f\in X^*, \,\|f\|=1\} \end{align}
MY TRIAL
It suffices to show that 
\begin{align} \forall\;\epsilon>0,\;\exists\;|f(x_{\epsilon})|\in \{|f(x)|:f\in X^*, \,\|f\|=1\}\;\;\text{such that}\end{align}
\begin{align} \|x\|-\epsilon< |f(x_{\epsilon})|\leq \|x\|.\end{align} 
Let $x\in X$ such that $x\neq 0.$ Otherwise, $\|f\|=0$. Then, by Hanh-Banach Theorem, there exists a linear functional $f$ on $X$ such that 
\begin{align} \|f\|=1 \;\;\;\text{and}\;\;\;|f(x)|= \|x\|\leq \|x\|.\end{align}
Please, I'm I right thus far? If yes, I am stuck here as I don't know how to arrive at 
\begin{align} \|x\|-\epsilon< |f(x)|.\end{align}
If no, can you help fix my wrong(s)?
 A: Credits to Martin Argerami. I present the full proof for future readers.
Let $x\in X$ be fixed, such that $x\neq 0.$ Consider the subspace spanned by $x$, defined by $Z=\{ \alpha x:\;\alpha\in \Bbb{R} \}.$
Define $f$ arbitrarily by
\begin{align} f:\,&Z\to \Bbb{R}\\& z\mapsto \alpha \| x \|.  \end{align}
Let $z\in Z$, then $|f(z)|= |\alpha \| x \||= \|\alpha x \|= \|z \|.\,$  This implies that $|f(z)|=\left|\|z \| \right|\leq \left|\|z \|\right|=\|z \|$ which gives boundedness of $f$ and $\|f \|=1.$
Let $\gamma,\lambda\in \Bbb{R}$ and $y,z\in Z,$ then $\exists\,\alpha,\beta \in \Bbb{R}$ such that $y=\alpha x$ and $z=\beta x.$ Then, $f$ is linear, since
\begin{align}f\left(\gamma y+\lambda z\right)&= f\left[(\gamma \alpha+\lambda \beta)x\right] \\&= (\gamma \alpha+\lambda \beta)\| x \|\\&= \gamma (\alpha\| x \|)+\lambda( \beta\| x \|)\\&= \gamma f\left( y\right)+\lambda f\left(z\right).\end{align}
Since $f$ is a linear functional, $\;\alpha f\left( x\right)=f\left(\alpha x\right)=f\left(z \right)=\alpha \| x \|$ implies that $f\left( x\right)=\| x \|.$ By the implication of $f$ being a bounded linear functional, we have by the Hanh-Banach Theorem,  that there exists $F\in X^{*}$ s.t. $F(x)=f(x)$ for all $x\in Z$ and 
\begin{align}F(x)=f(x)=\| x \|\;\;\text{and}\;\;\|F \|=\| f \|=1.\end{align}
So, 
\begin{align}\| x \|=\left|\| x \| \right|=\left| F(x) \right|\;\;\text{for some}\;\;F\in X^{*}\;\;\text{and}\;\;\|F \|=1.\end{align}
Taking $\sup $ over such $\;F\in X^{*}\;\text{and}\;\|F \|=1,$ we get 
\begin{align}\| x \|=\sup\{\left| F(x) \right|:\,F\in X^{*},\;\|F \|=1\}.\end{align}
A: Let’s denote $S_x =\sup\{|f(x)|:f\in X^*, \,\|f\|=1\}$.
We have to prove that $\Vert x \Vert =S_x$.
If $\Vert f \Vert =1$, then $\vert f(x)\vert \le \Vert x\Vert$. Hence $S_x \le \Vert x\Vert$. Conversely for $x\neq0$, on the subspace $\mathbb Rx$, the linear form defined by $f(y) = \lambda \Vert x \Vert$ for $y = \lambda x$ is such that $f(y) \le \Vert y \Vert$. Using Hahn Banach theorem we can extend $f$ to a linear form on $X$ such that $\vert f(y)\vert \le \Vert y \Vert$ for $y \in X$ and $\vert f(x) \vert = \Vert x \Vert$. Hence $S_x \ge \Vert x \Vert$, allowing us to conclude.
A: Your approach is flawed for two reasons that I see: first, you talk about an $x_\epsilon$, not sure what you expect by that; it's the $f$ that varies here, the $x$ is fixed. Second, you say 

"by the Hahn-Banach theorem, there exists a linear function $f$ on $X$ with $\|f\|=1$ and $|f(x)|=\|x\|$". 

That's not wrong: but it's precisely what you are supposed to prove. 
For any $f$ with $\|f\|\leq1$, you have $|f(x)|\leq\|f\|\,\|x\|=\|x\|$. So $|f(x)|\leq\|x\|$ for any $f$ in your set. And now, we use Hahn-Banach as you say. On the one-dimensional subspace $\mathbb C\,x$, define $f_0(\lambda x)=\lambda\,\|x\|$. Then 
$$
|f_0(\lambda x)|=|\lambda\,\|x\|\,|=\|\lambda x\|, 
$$
so $\|f_0\|=1$. Now, by Hahn-Banach, there exists $f\in X^*$, with $\|f\|=\|f_0\|=1$, and $f(x)=f_0(x)=\|x\|$. So 
$$
\|x\|=\max\{|f(x)|:\ f\in X^*,\ \|f\|=1\}. 
$$
