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!
I think that your first question is very broad and can have many interesting answers depending on the context. The point is that, since you are considering $C(X,Y)$, the answer depends a lot on what is the topology of the other space $Y$.
If you allow the topology of $Y$ to vary too, you can essentially make the set $C(X,Y)$ to be either the set of constant functions or the set of all functions. But if the topology of $Y$ is fixed (and somewhat nontrivial) you could have interesting behaviours, see the answer by Jochen.
Concerning the second question, as (again) Jochen suggests in a few comments, you can construct many examples, e.g. by composing any linear functional $f\in X^*$ with any continuous function $\phi:\mathbb R\to\mathbb R$.
In the above example, though, you are basically reducing yourself to the finite-dimensional case, where weak and strong topologies coincide. In fact, you can view any non-trivial functional $\phi\in X^*$ as a linear projection onto a one-dimensional space. I think that constructing a "genuinely infinite-dimensional" example is a bit less trivial (if not impossible, I don't know).
Even though this is not completely related, I want to mention the case of convex functions: if you have a continuous convex functional $\phi:X\to\mathbb R$, this is authomatically lower-semicontinuous in the weak topology of $X$ (this is related to the fact that a strongly closed, convex subset of a Banach space $X$ is also weakly closed), but you cannot upgrade this statement to continuous functions. In fact (again, again), Jochen suggests the example of the norm $||\cdot\||$, which is convex and continuous, thus weakly lower-semicontinuous, but not weakly continuous (there exist sequences of unitary vectors that converge weakly to zero). If you want to look at some references, have a look at the book of H. Brezis "Functional Analysis, Sobolev Spaces and Partial Differential Equations".
Best Answer
One goes about understanding a topology by looking at the basis, the "simplest" open sets of which all other sets are composed as unions. For a metric space, the basis consists of the balls, of course. And what's even nicer about topological vector spaces is that you only have to study what the basis sets look like at the origin, since topological vector spaces are shift-invariant.
The norm-topology's basis at the origin consists of the balls $ B(0,r) $. Easy enough. What do basis sets look like in the weak topology? Take a bounded linear functional $ f \in X^* $, and then look at $ \{ x \in X : |f(x)| < r \} $. You can hopefully imagine, this is "the space that is sandwiched between two hyperplanes". Now, this isn't quite the basis. To get a basis (at the origin), you need to consider all possible intersections of these "sandwiches". But the important thing is, you only get to "sandwich" in a finite number of directions $ f_1, \ldots, f_N $. So, in the infinite dimensional case, there's always some direction that you fail to sandwich. Hence, these basis sets are not bounded. This is counterintuitive, because in $ \mathbb{R}^3 $ (for example), you obtain a bounded set once you "sandwich" along the x-axis, y-axis, and z-axis (or in 3 directions of your choice).
Question 1 Let's first get clear on why the norm-topology contains the weak topology. Let $ U $ be open in the weak topology. Fix a point $ x_0 \in U $. We can find a basis set $ V $ with $ x \in V \subseteq U $. Recall, by "basis set", I mean a finite number of sandwiches, but we can be rigorous about this: for some numbers $ m_1 < M_1, \ldots m_N < M_N $ and $ f_1, \ldots f_N \in X^* $, the basis set $ V $ can be expressed $$ V = \{ x \in X : m_1 < f_1(x) < M_1, \ldots, m_N < f_N(x) < M_N \} $$ Now, it's pretty straightforward to find a ball $ B(x_0, r_0) $ that fits inside of $ V $. Do you see how? It relies on the boundedness of hte functionals $ f_1, \ldots f_N $. Thus we have $$ x_0 \in B(x_0, r_0) \subseteq V \subseteq U $$ which, you'll recall, shows that $ U $ is open in the norm topology (make sure you see why).
Question 1, Part b You then asked why $ x \mapsto |f(x)| $ is norm-continuous. But this is a composition of two continuous functions, the absolute value function and $ f \in X^* $ (which is continuous by definition).
Question 2 You asked about showing that $ g : x \mapsto \|Lx\| $ is weakly continuous, where $ L:X \rightarrow X $ is a bounded linear map. Surprisingly, it is not continuous with respect to the weak topology (in general). Indeed, pick $ X $ to be your favorite infinite-dimensional space, let's say $ X = \mathcal{l}^1(\mathbb{N}) $ and $ L = Identity : X \rightarrow X $. Then I claim $ g : x \mapsto \|Lx\| $ is not weakly continuous, since indeed $ g^{-1}((-a,a)) = B(0,a) $ is bounded, whereas weakly open sets in $ \mathcal{l}^1 $ are not bounded (!).
Addendum Fundamentally, the thing that's surprising here is that the norm $ x \mapsto \|x\| $ is not weakly continuous. To try to wrap our heads around this, consider the weak basis we discussed above, composed of finite intersections of "hyperplane sandwiches." This basis of the weak topology tells us that, in order for a function $ f : X \rightarrow \mathbb{R} $ to be weakly continuous, it is only allowed to change in a finite number of directions. The norm function, on the other hand, changes in all directions that you walk away from the origin.