In the general context of topological vector spaces (TVSs), those are vector spaces with a topology that makes the sum and product by scalars continuous (not necessarilly Hausdorff), I read the following lemma which I cannot prove:
Every bounded subset of a finite dimensional TVS is precompact.
In this paper, a subset of a TVS $A\subset X$ is bounded if for each $U\subset X$ neighborhood of $0$ there exists $\rho>0$ such that $A\subset \rho U$. On the other hand, $A$ is said to be precompact if for each such $U$, $A\subset A_0+U$, for some $A_0\subset A$ finite (where $+$ is the Minkowski sum).
I think I found a proof, thanks for the help. I'll try to make it as self contained as possible. Let $X$ be a finite dimensional vector space over the real or complex field, $A\subset X$ bounded in the sense above.
Consider $\overline{\{0\}}$ as a closed subspace of $X$, and the quotient space $X/\overline{\{0\}}=Y$. Then, $Y$ is a topological vector space with the topology that makes the canonical projection $\pi:X\rightarrow Y$ open, with dimension less than or equal to that of $X$, and is Hausdorff. Indeed, $0+\overline{\{0\}}\in Y$ is closed in $Y$ since its complement coincides with $\pi(X\setminus\overline{\{0\}})$, which is open in $Y$.
Since $A$ is bounded and $\pi$ is open, $\pi(A)$ is bounded. With the help of the Tychonoff theorem for topological vector spaces of finite dimension, we conclude that $\pi(A)$ is precompact in $Y$, as a subset of a compact set.
Let $U,V\subset X$ be neighborhoods of zero, with $U$ arbitrary and $V$ satisfying $V+V\subset U$. There exists $A_0\subset X$ finite such that $A+\overline{\{0\}}\subset A_0+V+\overline{\{0\}}$. From here, $A\subset A_0+V+\overline{\{0\}}\subset A_0+U$, and $A$ is precompact.