Why do we look at $X\rightarrow B$ transition in iodine? In the classic experiment where one investigates the absorption profile of $I_2$ vapor, we look at $X\rightarrow B$ electronic transitions where $X$ denotes the ground state and $B$ the second excited electronic state. Why is $X\rightarrow A$ not considered?

Note also that in the figure the eletronic state $A$ has the same limit $I+I$ (both atoms in the ground state) when the internuclear separation goes to infinity $r\rightarrow \infty$, wherears state $B$ has the limit $I+I^*$ indicating one excited atom. See also 2 who claims that $\Delta E(I,I^*)$ is the energy to excite a single Iodine atom. What is state $A$ then if not an electronic excitation?
 A: 
In the classic experiment where one investigates the absorption profile of $I_2$ vapor, we look at X→B electronic transitions where X denotes the ground state and B the second excited electronic state. Why is X→A not seen?

Presumably, if $X\to A$ is not seen, it is due to the molecular selection rules for photon absorption. (See the section titled "In Molecules" via the link in the previous sentence.)
The symmetry-based selection rules demand that the parity of the initial and final states are different. In particular, for a homonuclear molecular like $I_2$ the parity selection rules demand that (quotation from above-linked article):

"For homonuclear molecules, the g  ↔  u transition is allowed."

Based on the references provided by OP in the comments to OP's question, we also know that the ground state $X$ has molecular term symbol ${}^1 \Sigma_g$ (i.e., it has "g" or "even" parity). The $B$ excited state has molecular term symbol ${}^3 \Pi_u$ (i.e., it has "u" or "odd" parity).
The $X\to B$ transition is allowed by the parity selection rules, as well as the other relevant selection rules provided in the linked article. Therefore, we would expect to see this transition.
Presumable, the $A$ excited state electronic transistion is not allowed because either it has the wrong parity to allow a transition or it has other symmetry properties that disallow the transition (i.e., force matrix elements to be zero). OP may provide an updated reference for the $A$ state molecular term symbol to confirm this presumption, but has not done so yet.

By the way, what do I mean by "matrix elements" and where do these selection rules come from?
The matrix element that is usually considered when calculating absorption is:
$$
\langle F|\sum_i \vec r_i |I\rangle\;,
$$
where the $\vec r_i$ is the position operator of the $i$th electron, and the sum is over all the electrons in the molecule.
Because $\sum_i \vec r_i$ has odd parity, it can have no matrix elements connecting states of the same parity. This is where the parity selection rule and the other selection rules originate from.
Additional information on quantum mechanical calculations of absorption and aspects of symmetry can be found in various quantum mechanics textbooks and references therein. For example, Chapter 11 of the textbook by Bethe and Jackiv discusses selections rules.

Update (OP has updated the question)

Why is $X\rightarrow A$ not considered?

I do not think it is really possible for anyone other than the author of the figure to say why $X\rightarrow A$ was not considered. Presumably, $X\rightarrow A$ is suppressed by some selection rules and perhaps only shows up at higher-order in the interaction.

Note also that in the figure the eletronic state $A$ has the same limit $I+I$ (both atoms in the ground state) when the internuclear separation goes to infinity $r\rightarrow \infty$, wherears state $B$ has the limit $I+I^*$ indicating one excited atom. See also 2 who claims that $\Delta E(I,I^*)$ is the energy to excite a single Iodine atom. What is state $A$ then if not an electronic excitation?

The state $A$ is an electronic excitation of the molecule, but not of two separated atoms. That is, the $A$ electronic excitation is (roughly speaking) a combination of two atomic ground state orbitals with some phase difference that gives the molecular orbital a higher energy than the ground state molecular orbital, the latter of which is also (roughly speaking) a combination of two atomic ground state orbitals with some phase difference.
The basic idea of "roughly" can be understood by considering a weak coupling $\delta$ between two states that would otherwise have the same energy $\epsilon$. Such a two-state system is described by a hamiltonian like:
$\newcommand{\amat}[4]{\left(\begin{matrix}#1 & #2 \\ #3 & #4 \end{matrix}\right)}$
$$
\amat{\epsilon}{\delta}{\delta}{\epsilon}\;,
$$
which has energy eigenvalues:
$$
E_{\pm} = \epsilon\pm\delta.
$$
In this rough analogy, the energy of state $A$ is $\epsilon+\delta$ and the energy of state $X$ is $\epsilon - \delta$. In this rough analogy, the $\delta$ term goes to zero as the atomic separation goes to infinity and the $\epsilon$ value is the ground state atomic energy.
Clearly, as $\delta \to 0$ the energies approach each other, just like the energy of $A$ approaches the energy of $X$ as the separation goes to infinity.
