If $F$ is a family of linear functionals on a vector space $X$ then it induces a weak topology $w_F$ on $X$. The dual space of $(X,w_F)$ is the linear span of $F$ in the algebraic dual of $X$. In particular $(X,s)' = (X,w)'$ always holds.
This fully answers the first bullet. It also shows that there is no example as in the second bullet and that the procedure in the third bullet stabilizes at $(X,w)$ after one step.
The point is that if $\varphi \colon X \to \mathbb{F}$ is $w_F$-continuous then $U = \{x \in X \mid \lvert \varphi(x)\rvert \lt 1\}$ is a $w_F$-open neighborhood of $0$. By definition of the weak topology, there are $f_1,\dots,f_n \in F$ and $\varepsilon \gt 0$ such that $V = \bigcap_{i=1}^n \left\{x \in X \mid \lvert f_i(x)\rvert \lt \varepsilon\right\} \subseteq U$ and in particular $\bigcap_{i=1}^n \ker f_i \subseteq \ker \varphi $. Therefore $\varphi$ is a linear combination of the $f_i$ (a proof of this last assertion is here).
You can find a detailed proof in pretty much every book on topological vector spaces. A good reference is chapter 3 of Rudin's Functional Analysis, see in particular Lemma 3.9 and Theorem 3.10 (the Hausdorff assumption is only there because Rudin assumes locally convex spaces to be Hausdorff, as many authors do).
In what follows, $ \mathbb{F} $ shall denote either $ \mathbb{R} $ or $ \mathbb{C} $. The standard Euclidean topology on $ \mathbb{F} $ shall be denoted by $ \tau_{\mathbb{F}} $.
It is important to know that the weak topology on a vector space $ V $ over $ \mathbb{F} $ can only be defined after we have already endowed $ V $ with a linear topology, i.e., a topology on $ V $ that makes vector addition and scalar multiplication continuous operations. Hence, when one speaks of ‘weak topology’, it is always with respect to an existing linear topology.
What Semi-Norms Give Us
Let $ V $ be an abstract vector space over $ \mathbb{F} $ and $ \mathcal{P} $ a family of semi-norms on $ V $. We can define a topology on $ V $ by declaring its base to consist of subsets of $ V $ of the form
$$
\{ v \in V ~|~ {p_{1}}(v - x) < \epsilon ~ \land ~ \ldots ~ \land ~ {p_{n}}(v - x) < \epsilon \},
$$
where
This topology is called the semi-norm topology corresponding to $ \mathcal{P} $, and it can be shown to be a linear topology on $ V $. When we equip $ V $ with a semi-norm topology, we call $ V $ a locally convex space.
As you can see, semi-norms (whether there is only one or there are many) allow us to create a linear topology for $ V $ when there might have been none before. Its construction in no way depends on the weak topology, which comes only after.
Constructing the Weak Topology from an Existing Linear Topology
The weak topology on a vector space over $ \mathbb{F} $ can only be defined after we have equipped the vector space with some linear topology, for example the locally convex topology that was defined in the previous section.
Suppose that $ V $ is a vector space equipped with a linear topology $ \tau $. Let $ V^{*} $ denote the set of $ \tau $-continuous linear functionals on $ V $, i.e., continuous linear functions $ \varphi: (V,\tau) \to (\mathbb{F},\tau_{\mathbb{F}}) $. The weak topology on $ V $ with respect to $ \tau $ is defined as the smallest topology $ \tau_{\text{wk}} $ on $ V $ such that the mapping $ \varphi: (V,\tau_{\text{wk}}) \to (\mathbb{F},\tau_{\mathbb{F}}) $ is still continuous for all $ \varphi \in V^{*} $.
We can find a sub-basis of $ \tau_{\text{wk}} $ that consists of subsets of $ V $ of the form $ {\varphi^{\leftarrow}}[U] $, where $ \varphi \in V^{*} $ and $ U \in \tau_{\mathbb{F}} $.
Note that $ \tau_{\text{wk}} $ is a weaker topology than $ \tau $, i.e., $ \tau_{\text{wk}} \subseteq \tau $. Note also that $ \tau_{\text{wk}} $ is a linear topology on $ V $.
Wait! The Weak Topology Is Actually a Semi-Norm Topology!
I mentioned in the previous section that the weak topology arrives only after we have a linear topology. Here comes the tricky part. We can actually construct the weak topology as the semi-norm topology corresponding to some family $ \mathcal{P} $ of semi-norms. However, $ \mathcal{P} $ is not a rabbit pulled out of an empty hat. To define $ \mathcal{P} $, we need to have a linear topology in the first place. Herein lies the source of your confusion:
The weak topology is indeed a semi-norm topology, but you cannot begin to describe the semi-norms until you have put a linear topology on the vector space.
As before, let $ V $ be a vector space over $ \mathbb{F} $ equipped with a linear topology $ \tau $. For each $ \varphi \in V^{*} $, define a semi-norm $ p_{\varphi}: V \to [0,\infty) $ as follows:
$$
\forall v \in V: \quad {p_{\varphi}}(v) \stackrel{\text{def}}{=} |\varphi(v)|.
$$
Then $ \tau_{\text{wk}} $ as constructed in the previous section is actually the semi-norm topology corresponding to the family $ \{ p_{\varphi} ~|~ \varphi \in V^{*} \} $.
Motivation for the Weak Topology
As we already have $ \tau $, then why is there a need to create $ \tau_{\text{wk}} $? Think of it this way. The original topology $ \tau $ may contain more open sets than is actually necessary to make each $ \varphi \in V^{*} $ continuous. We can thus afford to throw out some open sets from $ \tau $ without affecting the continuity of the $ \varphi $’s. By discarding these unnecessary open sets and making the topology smaller, we can somehow endow $ V $ with certain nice topological properties. For example, with less open sets, some subsets of $ V $ automatically become compact subsets.
To give a further illustration, let us consider Kakutani’s Theorem, which states that a Banach space is reflexive if and only if its closed unit ball is compact with respect to the weak topology. For infinite-dimensional Banach spaces, the closed unit ball is never compact with respect to the norm-topology, which is a consequence of Riesz’s Lemma. However, by switching to the weak topology, the compactness of the closed unit ball may be regained, which is certainly the case precisely when the Banach space is reflexive, according to Kakutani’s result. Therefore, the weak topology gives us a complete characterization of reflexivity, and this is clear evidence of its utility.
Conclusion — $ V $ By Design
We saw a naked abstract vector space $ V $ who needed a topological outfit. We covered her shame by giving her a linear topology. $ V $ begged us, “Please respect my semi-norms if you wish to topologize me!” After we had dressed her, $ V $ examined herself in the mirror and delightfully exclaimed, “I look good with all my linear functionals!” We looked at her from top to bottom, shook our heads and told her, “No. You still look kinda baggy. But don’t worry! You’ll get to keep your linear functionals.” After removing her excess open sets, all of us concluded that $ V $’s minimalistic weak topology was stunning!
Best Answer
I explained the basic theory of initial topologies in this answer.
Given a family of functions defined on a set, where the codomain have topologies, we give the set the smallest topology that makes all these functions continuous wrt the already given topologies on the codomains. I prove existence and unicity in that answer.
In your case we have the set $X^\ast$ and all point-evaluation functions $\phi_x$, where $x$ ranges over $X$, and so the common codomain is the field $\Bbb R$ (or $\Bbb C$ in the case of complex vector spaces), which has its standard topology.
If we take all continuous linear maps from $X^\ast$ (in the norm topology) to $\Bbb R$) we get the weak topology (instead of the weak-star topology) on $X^\ast$; the evaluations are but a small subset of these. We then get a topology in general between (as subsets) the weak-star topology and the norm topology.
This is all general topology theory, but in the case where all $\phi_x$ are linear (as is the case here) and the domain is a vector space, we get a vector space topology on the domain, so that $+$ and scalar multiplication are still continuous on it.
So as to your question 2.: it's true by definition. And we choose it because it's useful, e.g. because the closed unit ball is compact in this topology (which can have useful consequences). And as the norm topology is one of the topologies that makes all $\phi_x$ continuous by minimality $\mathcal{T}_{w\ast} \subseteq \mathcal{T}_{\textrm{strong}}$ and so the inclusion is continuous. QED.
From the general theory plus some reasoning it follows that the following sets form a base for the open sets of the weak-star topology:
$$B(f; \{x_1, x_2, \ldots,x _n\}, r):= \{g \in X^\ast \mid \forall 1 \le i \le n: |f(x_i) - g(x_i) | < r\}$$
where $f \in X^\ast$, $\{x_1, \ldots, x_n\}$ is a finite subset of $X$ and $r>0$.
This makes it quite concrete as a topology.