# Infinite dimensional normed space whose only closed proper subspaces are finite dimensional

Is there a normed (infinite-dimensional) space $$V$$ such that the only closed proper subspaces are the finite-dimensional ones?. Without the "closed" hypothesis then it is clearly false (assuming the Axiom of Choice). I think the polynomials in $$[0,1]$$ may what I'm looking because of the Stone-Weierstrass theorem. Does a positive answer still hold when only talking about Banach spaces?

No. On any nonzero normed space $$X$$, there is (by the Hahn-Banach theorem) a nonzero continuous linear functional $$f$$. Its nullspace $$Y := \{x \in X : f(x) = 0\}$$ is a closed linear subspace of $$X$$. $$Y$$ is infinite dimensional provided $$X$$ itself is infinite dimensional.
No. Regarding your example, the polynomials on $$[0,1]$$ have, among many, the infinite-dimensional closed subspace $$V_0=\{p:\ p(0)=0\}.$$
For the general case, take $$v_0\in V$$ with $$\|v_0\|=1$$. Define a linear functional on $$\mathbb C v$$ by $$\varphi(\lambda v)=\lambda$$. Clearly $$|\varphi(\lambda v_0)|=|\lambda|=\|\lambda v_0\|.$$ So Hahn-Banach applies and we can extend $$\varphi$$ to all of $$V$$, with $$\|\varphi\|=1$$. Now define $$P:V\to V$$ by $$Pv=\varphi(v)\,v_0.$$ Then $$P$$ is an idempotent, and $$(I-P)V$$ is a proper infinite-dimensional subspace of $$V$$. Moreover, $$\|Pv\|=|\varphi(v)|\leq\|v\|,$$ so $$P$$ is bounded. Now suppose that $$\{v_j\}\subset (I-P)v$$ and $$v_j\to v$$. Then $$(I-P)v=\lim_j(I-P)v_j=\lim_jv_j=v,$$ and then $$v\in (I-P)V$$. So $$(I-P)V$$ is closed.