I've shown that for $X\subset \mathbb R^n$ it holds that:
$X$ is compact iff $X$ is closed and bounded.
Now I would like to show that this equivalence holds for any set in an $n$-dimensional vectorspace $V$.
I was thinking of using an isomorphism between $V$ and $\mathbb R^n$, but I'm not sure how. Could someone help me with this? Should I work with the coordinate map?
Edit
Ok, I was helped in the chat. I'm a little bit too lazy to write it out formally, but the general idea for the $"\impliedby"$ direction is the following: Let $\{v_1,\dots,v_n\}$ be a basis for $V$. Consider the coordinate mapping $f\colon V\to\mathbb R^n\colon v=\alpha_1v_1+\dots+\alpha_nv_n\mapsto (a_1\cdots a_n)$. Choose a closed and bounded subset of $X$. We know that $f(X)\subset \mathbb R^n$ is closed and bounded,$^*$ and by Heine-Borel we know that $f(X)$ is compact. Now consider the inverse $f^{-1}\colon\mathbb R^n\to V$. We know that $f^{-1}$ is continuous, so $f^{-1}(f(X))=X$ is compact.
$^*$: Showing that $f(X)$ is bounded is easy. Consider the $\infty$-norm for both $V$ and $\mathbb R^n$. We know there exists $M>0$ such that $\Vert v\Vert<M$ for all $v\in X$. This means that for each $w\in f(X)$, it holds that $\Vert w\Vert<M$ too (if we consider $\{e_1,\dots,e_n\}$ as our basis for $\mathbb R^n$). Showing that $f(X)$ is closed is left for others to write out, if they wish.