Proof: If $Ax=b$ has more than one solution, it has infinitely many solutions This is a follow up question to a question I asked yesterday: Link
I want to prove the following statement:

Let $A$ be an arbitrary $n \times n$ matrix, and let $Ax=0$ have more
  than one solution then it follows that $Ax=b$ van be solved for every
  $b$

I didn't know how to do it so I looked up the answer and tried to understand the proof from there: Link
The beginning is pretty straigh forward:
Let $u$ and $v$ be two different solutions 
$$\implies Au=0 , Av=0 \implies Au=Av \iff A(u-v)=0$$
since $u \not=v$ it follows that $u-v$ is not the zero solution. So far so good. However I don't understand this next step:
For $k \in \Bbb N, u+k(u-v)$ is a solution to $Ax=b$
$$A(u+k(u-v))=Au+Ak(u-v)=\color{blue}{x+0}=\color{blue}{x}$$
First, why does  $k \space \space \text{have to be} \in \Bbb N$ Why can't $\space  k \in \Bbb R$? Second, how is author getting the answer I marked in blue. How does getting $\color{blue}{x}$ tell me anything about the number of solutions? 

The statement underlined in skin-tone is blatantly false. If $Ax=0$ has more than one solution, which implies: at least one  solution $k\ne0$, then ${\rm dim\>ker}(A)\geq1$. Therefore $V:={\rm im}(A)$ has dimension $$\dim (V)=n-\dim\ker(A)<n\ .$$ It follows that ${\mathbb R}^n\setminus V$ contains infinitely many vectors $b$, and for all these $b$ the system $Ax=b$ has no solution.
