Intuition - If $Ax = b$ has infinitely many solutions, why can't $Ax = c$ have only one solution? [Strang P165 3.4. 22] 
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 $\mathbf{x_0 + \color{green}{(x_1 - x_2)}}$? 
 A: When you know one solution of the system $Ax=b$, call it $x_0$ and all solutions of the homogeneous system $Ax=0$, then you know all solutions of $Ax=b$.
Indeed, if $Ay=0$, then $A(x_0+y)=Ax_0+Ay=b+0=b$.
Conversely, if $x_1$ is a solution of $Ax=b$, then
$$
Ax_1=b=Ax_0
$$
What can you do with this? Observe that $b$ can be removed and you can write
$$
Ax_1=Ax_0
$$
Therefore $A(x_1-x_0)=0$, telling you that $y=x_1-x_0$ is a solution of $Ax=0$. But then $x_1=x_0+y$ has the desired form.
In particular, if $Ax=b$ has infinitely many solutions, also $Ax=0$ has.
Think to the equation of a line, $y=mx+q$. If $(x_0,y_0)$ and $(x_1,y_1)$ are points of the line, then $(x_1-x_0,y_1-y_0)$ is a point on the line $y=mx$. Indeed
$$
y_1-y_0=(mx_0+q)-(mx_1+q)=m(x_1-x_0)
$$
The points of the line $y=mx+q$ can be obtained by translating all the points of $y=mx$ by the same point.
Now let's see your particular problem. There is another fact to keep in mind:

if a linear system has two distinct solutions, then it has infinitely many solutions

This is because only the following cases can happen for a system: it has


*

*no solution, or

*exactly one solution, or

*infinitely many solutions


For, assume $Ax=b$ has two distinct solutions $x_1$ and $x_2$; then, by the same reasoning as above, $y=x_1-x_2$ is a solution of $Ax=0$. But then also $\alpha y$ is a solution of $Ax=0$ for all scalars $\alpha$, so the system $Ax=b$ has infinitely many solutions, because every vector of the form $x_1+\alpha y$ is a solution.
You now should see that showing that a system has infinitely many solutions is equivalent to showing it has two distinct solutions.
Now, solvability of the system $Ax=c$ is equivalent to $c\in\operatorname{colspace}(A)$; so, assume $Ax=c$ is solvable and $Ax=b$ has infinitely many solutions. Then


*

*$Ax=b$ has infinitely many solutions, so

*$Ax=0$ has infinitely many solutions, so

*$Ax=c$ has infinitely many solutions.

A: 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 $T$.
