This isn't right. First, intuitively, if $\vec{k}$ is a least-squares solution, this means that $A\vec{k}$ is close to $\vec{b}$. Multiplying $\vec{k}$ by some arbitrary real number would multiply $A\vec{k}$ by the same number, and this could take it very far from $\vec{b}$.
To engage with your thinking more directly, if $\vec{p}$ is the orthogonal projection of $\vec{b}$ into the column space of $A$, then $c\vec{p}$ is not in general. Visualizing this in 3-space is helpful. Say $\vec{b}$ is a point in 3-space. If $A\vec{x}=\vec{b}$ doesn't have a solution, then the column space of $A$ is some subspace of 3-space that doesn't include $\vec{b}$. Say it is a plane. Then if $\vec{k}$ is the least squares solution, $\vec{p}=A\vec{k}$ is the unique point in this plane that is closest to $\vec{b}$. Scaling $\vec{p}$ by something will take it further away from $\vec{b}$.
EDIT: now that you have edited the question,
YES, you are right; if $\vec{k}$ is the least-squares solution of $A\vec{x}=\vec{b}$, then $c\vec{k}$ is the least-squares solution of $A\vec{x}=c\vec{b}$. Furthermore, your reasoning is correct, though imprecisely articulated (and $\vec{b}$ should be $c\vec{b}$ at the end of the paragraph). I take your question to be about how to make your proof more precise.
Here is one way to do it. As you note, saying that $\vec{k}$ is a least-squares solution of $A\vec{x}=\vec{b}$ is equivalent to saying that $A\vec{k}=\vec{p}$ is the orthogonal projection of $\vec{b}$ into the column space of $A$. Let $\vec{q}=\vec{b}-\vec{p}$, so that $\vec{b}=\vec{p}+\vec{q}$, $\vec{p}$ is in $A$'s column space, and $\vec{q}$ is orthogonal to every vector in $A$'s column space.
Multiplying everything by $c \in \mathbb{R}$, we have $c\vec{b}=c\vec{p}+c\vec{q}$; $c\vec{p}$ is still in $A$'s column space (you noted this in your informal proof), and $c\vec{q}$ is still orthogonal to everything in $A$'s column space, since as a multiple of $\vec{q}$, it is orthogonal to everything $\vec{q}$ is orthogonal to. (This is what was missing from your argument, although you intuited it). This means that $c\vec{p}$ is the orthogonal projection of $c\vec{b}$ into $A$'s column space. Since $A(c\vec{k})=cA\vec{k}=c\vec{p}$ as you note, this means that $c\vec{k}$ is a least-squares solution to $A\vec{x}=c\vec{b}$.