This might be obvious, but I cannot prove it right away nor find a reference.
Consider a matrix $A\in\mathbb{R}^{m\times n}$, $m>n$ having full column rank and a set $X$ of vectors $x\in\mathbb{R}^{n}$.
We know that the set $Y$ of vectors $y=Ax$ is bounded, that is, for every $x\in X$, $\lVert A x \rVert<y_{max}$ for some $y_{max}\in\mathbb{R}^+$ and some vector norm $\lVert \bullet \rVert$.
Does this imply then that $X$ is itself bounded, that is, there is $x_{max}\in\mathbb{R}^+$ such that $\lVert x \rVert<x_{max} \;\forall\; x \in X$?
Since we are working on finite dimensional spaces, all norms are equivalent Hence I'll don't mention that.
Note that the hypotheses imply that $A$ defines an injective linear map $A\colon \mathbb{R}^n \longrightarrow \mathbb{R}^m$. In this case, $A$ is an embedding of $\mathbb{R}^n$ into $\mathbb{R}^m$, in the topology defined by the norms. In particular, $A$ is proper. On the other hand $\overline{Y}$ is compact by hypothesis. Then we have $$ X \subset A^{-1}(\overline{Y}) $$ $A$ being proper implies $A^{-1}(\overline{Y})$ compact which implies $X$ bounded.