This argument avoids the use of ultrafilters.
Let $e_n \in \ell_\infty^*$ be the evaluation map $e_n(x) = x_n$. The set $\{e_n\}$ is contained in the unit ball of $\ell_\infty^*$, which by Alaoglu's theorem is weak-* compact, hence $\{e_n\}$ has a weak-* cluster point; call it $f$. I claim this $f$ has the properties you desire.
It is easy to check that the set of multiplicative linear functionals is weak-* closed in $\ell_\infty^*$; each $e_n$ is multiplicative and hence so is $f$.
For each $x \in \ell^\infty$, let $\pi_x : \ell_\infty^* \to \mathbb{C}$ be the evaluation functional $\pi_x(g) = g(x)$. By definition of the weak-* topology, $\pi_x$ is weak-* continuous. Suppose $x \in c \subset \ell_\infty$ is convergent, with $x_n \to a$. By continuity of $\pi_x$, $f(x) = \pi_x(f)$ must be a cluster point of $\{\pi_x(e_n)\} = \{x_n\}$. But $x$ is a convergent sequence so the only cluster point of $\{x_n\}$ is $a$. Thus $f(x) = a$.
$f$ has another interesting property: since $\mathbb{C}$ is metric, all cluster points in $\mathbb{C}$ are subsequential limits. Thus for any $x \in \ell^\infty$, $f(x)$ is a subsequential limit of $\{x_n\}$. For instance, if $x_n = (-1)^n$, $f(x)$ must be either -1 or 1, whereas a Banach limit must assign $x$ the limit value of 0. A corollary of this is that for any real sequence $\{x_n\}$, we must have $\liminf x_n \le f(x) \le \limsup x_n$.
It is also interesting to note that $f$ is an element of $\ell_\infty^*$ that cannot correspond to an element of $\ell^1$. So this gives an alternate proof that $\ell^1$ is not reflexive.
If you like, you can instead produce $f$ as a limit of a subnet of $\{e_n\}$, or a limit of $\{e_n\}$ along an ultrafilter. In any case it is a cluster point.
First, the following easy corollary of Hahn-Banach for normed spaces might be useful:
Proposition: Let $X$ be a normed space, and let $Y$ be a subspace with $\phi\in Y^*$. Then $\phi$ has an extension to $\tilde{\phi}\in X^*$ such that $\|\phi\|=\|\tilde{\phi}\|$.
To prove this, apply the usual Hahn Banach theorem with $p(x)=\|\phi\|\|x\|$
Now, as you did before, define
$\psi_0:W\to\mathbb{R}$ by
$\psi_0((x_n)_n)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}x_n$
By Hahn-Banach, we can extend this to a function $Lim:l^{\infty}\to\mathbb{R}$ such that
$\|Lim\|=\|\psi_0\|$
Let's prove part c) first now. In light of the above, we need only show that
$\|\psi_0\|=1$. Indeed, we have $\frac{1}{n}\sum_{k=1}^{n}x_n\leq \frac{n\|(x_n)_n\|_{\infty}}{n}=\|(x_n)_n\|_{\infty}$
Thus $\|\psi_0\|\leq 1$. To see that $\|\psi_o\|=1$, consider the sequence $x\in W$ with only ones as its entries.
This proves c).
For part a)
Note that by linearity:
$Lim(x_1,x_2,...)=Lim(x_1-x_2,x_2-x_3,x_3-x_4,...)+Lim(x_2,x_3,x_4,...)$
$=\psi_0(x_1-x_2,x_2-x_3,x_3-x_4,...)+Lim(x_2,x_3,x_4,...)=Lim(x_2,x_3,x_4,...)$
This proves a)
Finally, for part b), let $x=(x_n)\in l^{\infty}$ Choose $a,b$ so that
$a<\liminf_{n\to\infty}(x_n)$ and
$b>\limsup_{n\to\infty}(x_n)$
Then there is $N$ such that for $n>N$, we have $a<x_n<b$.
Denote now 1 by the sequence with only ones as its entries and denote $y=(y_n)_n$ as the sequence such that $y_n=x_{n+N}$
Then $w:=y-a\cdot$1 and $z:=b\cdot$1$-y$ are positive and bounded sequences.
Thus,
$0\leq Lim(w)=Lim(y)-a=Lim(x)-a$
So that, $a\leq Lim(x)$
where the last equality was obtained using the shift invariance proven in a)
Similarly, we obtain $Lim(x)\leq b$.
Since, $a$ and $b$ were arbitrary, b) follows.
The one step that needs justification is that if $x=(x_n)_n$ is positive and bounded, then $Lim(x)\geq 0$
Indeed, without loss of generality, $\|x\|\leq1$(otherwise, rescale x)
Now, $1-Lim(x)=Lim($1$-x)\leq\|$1$-x\|\leq 1$
Best Answer
Let $a=\liminf x_n.$ For $\varepsilon >0$ there is $N$ such that $x_n>a-\varepsilon$ for $n>N.$ Define $$y_n=\begin{cases} a-\varepsilon & n\le N\\ x_n & n>N \end{cases}$$ Then $$L(y)=L(x)+L(y-x)=L(x)+\lim_n(y_n-x_n)=L(x)$$ as the sequence $y-x$ vanishes for $n>N.$ We have $y_n-(a-\varepsilon)\ge 0.$ Therefore, by applying $(1)$, we get $L(y-(a-\varepsilon){\bf 1})\ge 0,$ where $\bf 1$ denotes the sequence with all terms equal $1.$ This implies $$L(y)\ge (a-\varepsilon)L({\bf 1})=(a-\varepsilon)\lim_n 1=a-\varepsilon $$ As $\varepsilon >0$ is arbitrary we get $L(x)\ge a.$
A similar reasoning is valid for the other inequality, or we can use the trick replacing the sequence $x_n$ with $-x_n.$