Suppose $f:U\subseteq\mathbb{R^n}\to\mathbb{R^n}$ is of class $C^1$ and for some $a\in U$, $Df(a)$ is a linear isomorphism.
Prove there exists a $V\subseteq U$ such that it contains $a$ and also for all $p\in V$, $Df(p)$ is a linear isomorphism.
I know it has something to do with the inverse function theorem, but I can not figure out a way to prove it.
I don't think you need the inverse function theorem here. All you need to know is that $Df(p)$ is a linear isomorphism at a given $p \in U$ if and only if $\det Df(p) \neq 0$. The mapping $p \mapsto Df(p)$ is continuous (because $f$ is continuously differentiable) and the mapping $A \mapsto \det A$ is continuous, so the composition $p \mapsto \det Df(p)$ is continuous. Therefore, the set $\{ p \in U : \det Df(p) \neq 0 \}$ is open, and your result follows.
Having said that, the official statement of the inverse function theorem does tell you that $Df(p)$ is invertible for $p$ in a local neighbourhood of $a$, which is precisely the result you want. It even gives you a formula for the inverse $Df(p)^{-1}$. But quoting the inverse function theorem here seems overkill, because the non-trivial part of the inverse function theorem is really that the function $f$ itself has a local inverse.