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?
Thanks in advance
On
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.
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.