Does the following proof work?
Here is the statement I am trying to prove: Let $(V,\lVert\cdot\rVert)$ be a finite-dimensional normed vector space, and let $T:V\to V$ be a distance-preserving linear map. Show that $T$ is a bijection.
Proof: Let $e_1,\ldots,e_n$ be a basis for $V$. Notice that $T$ is bijective iff $T(e_1),\ldots,T(e_n)$ is a basis for $V$ iff $T(e_1),\ldots,T(e_n)$ are linearly independent. Assume that they are not. Then, relabeling if necessary, we can write $$T(e_n)=c_1T(e_1)+\dots+c_{n-1}T(e_{n-1})$$ for some scalars $c_1,\ldots,c_n$. But then, we find that $$0=\lVert c_1T(e_1)+\dots+c_{n-1}T(e_{n-1})-T(e_n)\rVert = \lVert T(c_1e_1 +\dots c_{n-1}e_{n-1}-e_n)\rVert=\lVert c_1e_1 +\dots c_{n-1}e_{n-1}-e_n\rVert \neq 0$$ which is a contradiction. Therefore, $T$ must be a bijection. This concludes the proof.
I'd also be interested to know whether or not this statement holds if the assumption the $V$ is finite-dimensional is removed, and if you know of any proofs of this statement that do not explicitly use a basis for $V$.
I think your proof works.
You can also prove this result directly without explicitly using a basis for $V$. Let $v \in V$ be such that $T(v) = 0$. Since $T$ is distance-preserving, $0 = \Vert T(v) \Vert = \Vert v \Vert$, which shows that the kernel of $T$ is trivial. So $T$ is injective. By the rank theorem, the dimension of $V$ is equal to the rank of $T$, which implies that $T$ is also surjective and hence bijective.
If the assumption that $V$ is finite-dimensional is removed, then I don't think your statement holds. See this post for example: Not bijective isometry