A vector space is finite dimensional if all of its proper subspaces are finite dimensional

You need to use the fact that every linearly independent set can be extended to a basis.

Then you can take any nonzero vector $v$, extend $\{v\}$ to a basis $B$, and consider the span of $B\setminus\{v\}$. Being finitely dimensional, it means that $B\setminus\{v\}$ is finite, so $B$ is finite, and so $V$ has a finite dimension.

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It might be relevant to point out that the fact "every linearly independent set can be extended to a basis" is equivalent to the axiom of choice. Of course we need only a small fraction of choice for this specific proof (although that would alter the formulation a bit).

Nevertheless, it is consistent without the axiom of choice that there is a vector space which is not finitely dimensional, but every proper subspace does in fact have a finite dimension. Weird.

Let $V$ be an infinite dimensional vector space, let $v$ be a vector of $V$. Then in particular there exists a countable subset $\{v_1,\cdots, v_n,\cdots\}$ of $V$ such that $\{v_1,\cdots, v_n,\cdots\}$ spans a subspace of $V$. We may assume $\{v_1,\cdots, v_n,\cdots\}$ is a basis, so in particular $\{v_2,\cdots, v_n,\cdots\}$ spanns a proper subspace $U$ of $V$ which is of countable dimensionality. The finite dimensional case is automatic.

Your proof for inner product spaces can be generalized (in some sense) in the following manner:

Let $U$ be a (nonzero) subspace of $V$, and let $\ \mathcal U=\{u_1,\ldots, u_n\}$ be a basis for $U$. Using Zorn's lemma we can extend $\ \mathcal U$ to a basis $\mathcal V$ of $V$. Let $U'$ be the subspace spanned by $\mathcal V \setminus \mathcal U$. Then $V=U\oplus U'$, and since both $U$ and $U'$ are finite-dimensional it follows that $V$ is finite-dimensional.