Proof that $l^1(X)^*$ is indeed $l^\infty (X)$ I know this is kind of well-known and is usually given as an example. But
how does one actually prove it?
Let's assume that $ X$ is a countable set with discrete topology (for simplicity's sake).
Now
$$l^1(X)=\{f:X\to \mathbb{C}\ \ |  \|f\|_1=\sum_{x\in X}|f(x)|<\infty\ \}$$
and
$$l^\infty(X)=\{f:X\to \mathbb{C}\ \ | \|f\|_\infty= \sup_{x\in X}|f(x)|<\infty\ \}$$
Now I would like to prove that $l^1(X)^*$ is indeed isomorphic (I presume as Banach spaces?) to $l^\infty(X)$.
$l^1(X)^*$ is the set of all bounded linear functionals, meaning
$$l^1(X)^*=\{\omega:l^1(X)\to \mathbb{C}\ \ | \|\omega\|<\infty \ \} $$
Where $$\|\omega\|=\sup_{f\in l^1}\dfrac{|\omega(f)|}{\|f\|_1}$$
Now I would like to show that if $\|\omega\|<\infty$ there is a one-to-one correspondence with some $g\in l^\infty(X)$.
I was thinking of the following
Take the canonical (Schauder) basis of $l^1(X)$; namely $\{\delta_x\}_{x\in X}$ and write
$$f=\sum_{x\in X}\alpha_x \delta_x$$
Then the condition for $\|f\|_1<\infty$ is equal to $\sum_x |\alpha_x|<\infty$
By linearity, I can also write the norm of some $\omega$ as (here I think of $(\alpha_x)$ as some sequence of numbers.
$$\|\omega\|=\sup_{\alpha_x}\dfrac{\sum_x |\alpha_x| |\omega(\delta_x)|}{\sum_x |\alpha_x|}$$
But here I get stuck. What can I do, or is there a better approach?
 A: You have a good starting point, but of course the goal is to find the function $g$ that will correspond to your arbitrary $\omega$!
I'll use $\ell^1 = l^1(\mathbb{N})$ and $\ell^\infty = l^\infty(\mathbb{N})$ to denote these two spaces.
There is a map $\Phi : \ell^\infty \to (\ell^1)^*$ given by $\Phi(f)(g) = \int fg$, or in other words
$$\Phi(f)(g) = \sum_{n=1}^\infty f(n) g(n).$$
This function is clearly linear. Next, we will show that $\Phi$ is an isometry.
Let $f \in \ell^\infty$ be arbitrary. Then, for any $g \in \ell^1$, we have
$$\lvert \Phi(f)(g) \rvert = \left\lvert \sum_{n=1}^\infty f(n)g(n) \right\rvert \leq \sum_{n=1}^\infty \lvert f(n) \rvert \lvert g(n) \rvert \leq \sum_{n=1}^\infty \lVert f \rVert_\infty \lvert g(n) \rvert = \lVert f \rVert_\infty \sum_{n=1}^\infty \lvert g(n) \rvert = \lVert f \rVert_\infty \lVert g \rVert_1.$$
Thus, $\lVert \Phi(f) \rVert \leq \lVert f \rVert_\infty$. Next, let $\varepsilon > 0$ be arbitrary, and let $x \in \mathbb{N}$ be such that
$$\lVert f \rVert_\infty - \lvert f(x) \rvert < \varepsilon.$$
Letting $\delta_x \in \ell^1$ denote the indicator function of $x$, we see that
$$\lvert \Phi(f)(\delta_x) \rvert = \lvert f(x) \rvert > \lVert f \rVert_\infty - \varepsilon.$$
Since $\lVert \delta_x \rVert_1 = 1$, this implies that $\lVert \Phi(f) \rVert \geq \lVert f \rVert_\infty - \varepsilon$. Since $\varepsilon$ was arbitrary, this implies that $\lVert \Phi(f) \rVert \geq \lVert f \rVert_\infty$. We conclude that $\lVert \Phi(f) \rVert = \lVert f \rVert_\infty$. Since $f$ was arbitrary, this says that $\Phi$ is an isometry, as desired.
We only have left to show that $\Phi$ is surjective. So, let $\psi \in (\ell^1)^*$ be arbitrary. We hope that $\psi = \Phi(f)$ for some $f \in \ell^\infty$. If that were to happen, then for all $n \in \mathbb{N}$, we would have
$$\psi(\delta_n) = \Phi(f)(\delta_n) = f(n).$$
Thus, if any function $f$ is to work, it must be given by $f(n) = \psi(\delta_n)$.
So, we now define $f : \mathbb{N} \to \mathbb{C}$ by $f(n) = \psi(\delta_n)$. We see that, for all $n \in \mathbb{N}$, we have
$$\lvert f(n) \rvert = \lvert \psi(\delta_n) \rvert \leq \lVert \psi \rVert \lVert \delta_n \rVert_1 = \lVert \psi \rVert.$$
Thus, $f \in \ell^\infty$. Finally, we claim that $\Phi(f) = \psi$. Let $g \in \ell^1$ be arbitrary. For each $N \in \mathbb{N}$, define $g_N : \mathbb{N} \to \mathbb{C}$ by
$$g_N = \sum_{n=1}^N g(n) \delta_n,$$
so that each $g_N$ is finite sum of $\delta_n$'s and $g_N \to g$ pointwise. Then by dominated convergence we get that $g_N \to g$ in $\ell^1$, so by continuity we have $\lim_{N \to \infty} \psi(g_N) = \psi(g)$. Also, by linearity we have
$$\lim_{N \to \infty} \psi(g_N) = \lim_{N \to \infty} \psi \left(\sum_{n=1}^N g(n) \delta_n\right) = \lim_{N \to \infty} \sum_{n=1}^N g(n) \psi(\delta_n) = \sum_{n=1}^\infty g(n) f(n) = \Phi(f)(g).$$
Since $g$ was arbitrary, we conclude that $\psi = \Phi(f)$, as desired.
We have now shown that $\Phi$ is a surjective isometry, and thus $\Phi$ is an isometric isomorphism. $\square$
