Consider the sequence $\left\{e_n\right\}_{n \in \mathbb{N}} \subset \ell^p(\mathbb{F}), 1 \leq p \leq \infty$, of unit vectors. Then $\left\|e_n\right\|_p=1$ for all $n \in \mathbb{N}$, and $\left\{e_n\right\}_{n \in \mathbb{N}}$ is not a Cauchy sequence for any $p$.
For $1<p<\infty$, we can now use the representation $\ell^p(\mathbb{F})^* \cong \ell^q(\mathbb{F})$ : for every $x^* \in$ $\ell^p(\mathbb{F})^*$, there exists a $y \in \ell^q(\mathbb{F})$ such that $$ \left\langle x^*, e_n\right\rangle_{\ell^p}=\sum_{k=1}^{\infty} y_k\left(e_n\right)_k=y_n \rightarrow 0, $$ since $\|y\|_q<\infty$ implies that $\left\{y_k\right\}_{k \in \mathbb{N}}$ is a null sequence. Hence $e_n \rightharpoonup 0$ in $\ell^p(\mathbb{F})$ for $1<$ $p<\infty$, and therefore also $e_n \rightharpoonup^* 0$ by reflexivity ($X$ is reflexive, then weak convergence in $X$ and weak-* convergence in $X^**$ coincide)
Since $\ell^1(\mathbb{F})$ is not reflexive, we have a choice here: if we consider $e_n \in \ell^1(\mathbb{F})$ via the isomorphism $T:\ell^1(\mathbb{F})\to c_0(\mathbb{F})^*$, we have $$ \left\langle T e_n, y\right\rangle_{c_0}=\sum_{k=1}^{\infty}\left(e_n\right)_k y_k=y_n \rightarrow 0 \quad \text { for all } y \in c_0(\mathbb{F}) $$ and hence $e_n \rightharpoonup^* 0$ in $\ell^1(\mathbb{F})$.
On the other hand, we do not have $e_n \rightharpoonup 0$, since the constant sequence $y:=\{1\}_{k \in \mathbb{N}} \in \ell^{\infty}(\mathbb{F}) \cong \ell^1(\mathbb{F})^*$ satisfies $$ \left\langle T y, e_n\right\rangle_{\ell^1}=\sum_{k=1}^{\infty}\left(e_n\right)_k y_k=y_n=1\neq 0 \quad \text { for all } n \in \mathbb{N} . $$
But $c_0(\mathbb{F}) \subset \ell^{\infty}(\mathbb{F})$, and hence the only candidate for the weak limit is 0. This shows that $\left\{e_n\right\}_{n \in \mathbb{N}}$ cannot converge weakly at all.
My questions regarding this example are:
Why does $c_0(\mathbb{F}) \subset \ell^{\infty}(\mathbb{F})$ mean that the only candidate for the weak limit is 0? I dont get the relation of what was proven so far with this inclusion.
Only the first part of the example (where $1 < p < \infty$) shows that weak convergence does not imply strong convergence, right? What conclusion can I take from the second part where $p=1$?
Assume that $e_n \rightharpoonup u$ in $\ell^1$ where $u\in\ell^1$.
Case 1: $u\ne 0$. Then there exists some $n_0$ such that $u_{n_0}\ne 0$. Let $F_1$ be the functional on $\ell^1$ given by $F_1(v)=v_{n_0}$. It is bounded because $|v_{n_0}|\le\|v\|_1$ for all $v\in \ell^1$. We have $F_1(e_n)=0$ for all $n>n_0$ so $\lim_{n\to+\infty}F_1(e_n)=0$. On the other hand $F_1(u)=u_{n_0}\ne 0$. This contradicts $e_n \rightharpoonup u$.
Case 2: $u=0$. Let $F_2$ be the functional on $\ell^1$ given by $F_2(v)=\sum_{k=1}^\infty v_k$. This is also bounded because $|\sum_{k=1}^\infty v_k|\le \sum_{k=1}^\infty |v_k|=\|v\|_1$. We have $F_2(e_n)=1$ for all $n$, but $F_2(u)=F_2(0)=0$, contradicting $e_n \rightharpoonup u$.
In either case there's a contradiction to $e_n \rightharpoonup u$ so the sequence $(e_n)_n$ does not converge in $\ell^1$.
Schur's theorem (from 1921) proves that weak convergence of a sequence in $\ell^1$ is equivalent to strong convergence of the same sequence.
This does not imply that the weak topology and strong topology on $\ell^1$ coincide. A topology is not always characterized by convergence of sequences. The norm topology on $\ell^1$ is metric, and the topology of a metric space is characterized by sequences. The weak topology of a Banach space does not have to be metrizable.
The weak topology of $\ell^1$ is not identical to the strong topology, because every basis element defining a weak open neighborhood of zero contains a subspace of $\ell^1$ of finite co-dimension, while $\ell^1$ is infinite-dimensional. The norm-unit-ball of $\ell^1$ does not contain any non-zero subspace. As a consequence of Schur's theorem we get that the weak topology of $\ell^1$ is not metrizable.