A corollary of Hahn–Banach theorem states that
Let $E$ be a normed vector space, $M$ a proper closed subspace and $x \in E$. If $d(x,M) = \delta > 0$, so exists $f \in E'$ such that $\|f\|=1$, $f(x)=\delta$ and $f(m)=0$ $\forall m \in M $.
Consider $T: \ell_\infty \rightarrow \ell_\infty$ a bounded linear operator defined by $$T(x_1,x_2,x_3,\dots) = (x_2,x_3,\dots).$$
Let $M=\{ x-T(x) : x \in \ell_\infty\}$, so $M$ is a subspace of $\ell_\infty$. If $e=(1,1,1,\dots) \in \ell_\infty$, so $d(e,M)=1>0$. Then, applying the corollary above in $\overline{M} $, exists $f \in \ell_\infty'$ such that $\|f\|=1$, $f(e)=d(x, \overline{M}) = d(e, M) =1$ and $f(x)=0~\forall x \in M\subset \overline{M} $.
I was able to show that $$f(x_1,x_2,x_3,\dots) = f(x_2,x_3,\dots) ~~\forall (x_n) \in \ell_\infty.$$ Besides that, we have that $\forall x = (x_n) \in c = \{ (x_n) \in \ell_\infty : x_n \text{ is convergent} \}$ $$f(x) = \lim_{n \rightarrow \infty} x_n$$
Now, let $x=(x_n), y=(y_n) \in \ell_\infty$ such that $x_n \geq y_n$ $\forall n \in \mathbb{N}$. How can I show that $f(x) \geq f(y)$?