If $\mathbf{Ax = b}$ has infinitely many solutions, why is it impossible for $\mathbf{Ax = c}$
(where $\mathbf{c}$ is a new right side) to have only one solution?
Proof : Take two solutions of $\mathbf{Ax = b} :$ $\mathbf{Ax_1 = b}$ and $\mathbf{Ax_2 = b} \implies \mathbf{\color{green}{A(x_1 - x_2) = 0}}$.
Thus, if $\mathbf{Ax_0 = c}$ then add the homogenous solution in green to $\mathbf{x_0}$: $\mathbf{Ax_0 + \color{green}{A(x_1 - x_2)} = c + \color{green}{0} }$ so $\mathbf{x_0}$ is not unique.
If $\mathbf{c}$ is not in $colspace(A)$, then no solution to $\mathbf{Ax = c}$.
$\Large{{1.}}$ I accept the proof but I don't perceive the intuition? Would someone please explain/uncloak it?
I recollect: If $\mathbf{c}$ is not in $colspace(A)$, then no solution to $\mathbf{Ax = c}$.
$\Large{{2.}}$ How would you divine/previse the strategy of subtracting two solutions to $\mathbf{Ax = b} $ and then adding this homogeneous solution to $\mathbf{Ax = c}$ ?
$\Large{{3.}}$ Does the above prove that $\mathbf{Ax = c}$ has infinitely many solutions also,
beyond $\mathbf{x_0}$ and $\mathbf{x_0 + \color{green}{(x_1 - x_2)}}$?
This is mostly a response to "1.", but indirectly addresses "2." and "3." as well.
If $V$ and $W$ are arbitrary vector spaces, the level sets of a linear transformation $T:V \to W$, a.k.a., the solution sets of $T(\mathbf{x}) = \mathbf{b}$ as $\mathbf{b}$ ranges over the image $T(V)$, are affine subspaces of $V$; specifically, they're translates of $\ker(T)$. Any two translates of $\ker(T)$ are clearly in bijective correspondence. (This observation is nothing but a geometric reformulation of your algebraic argument, but perhaps that's the intuition you're seeking.)
A bit more algebraically, if you regard $V$ and $W$ as Abelian groups under vector addition, then a linear transformation $T$ induces a group homomorphism, and the level sets of a homomorphism are left cosets (i.e., translates) of the kernel, since $T(\mathbf{x}_1) = T(\mathbf{x}_2)$ if and only if $T(\mathbf{x}_1 - \mathbf{x}_2) = \mathbf{0}$, if and only if $\mathbf{x}_1 - \mathbf{x}_2 \in \ker(T)$, if and only if $\mathbf{x}_1 = \mathbf{x}_0 + \mathbf{x}_2$ for some $\mathbf{x}_0$ in $\ker(T)$.
Finally, in case a visual example helps: Fix real numbers $a$ and $b$, not both $0$, and define $T:\mathbf{R}^2 \to \mathbf{R}$ by $T(x, y) = ax + by$. The level sets of $T$ are the parallel lines $ax + by = c$. Again, this picture is essentially general: $V$ is partitioned into parallel affine subspaces, i.e., non-empty level sets of $T$.