Is every finite codimensional subspace of a Banach space closed? Is it also complemented? I know how to answer the same questions for finite dimensional subspaces, but couldn't figure out the finite codimension case.

It's a standard result that a linear functional from a Banach space to the underlying field (real or complex numbers) is continuous if and only if the its kernel is closed. Notice that its kernel is of codimension one. So, use the axiom of choice to find a discontinuous linear functional, and you have found a codimension one subspace which isn't closed. (As I was typing this, rpotrie got the same answer...) As for complementation: well, this only makes sense for closed finite codimension subspaces. But then it's a perfectly reasonable question, and the answer is "yes". If F is of finite codimension in E, then by definition we can find a basis $\{x_1,\cdots,x_n\}$ for E/F. For each $k$ let $x_k^*$ be the linear functional on $E/F$ dual to $x_k$, so $x_k^*(x_j) = \delta_{jk}$. Then let $\mu_k$ be the composition of $E \rightarrow E/F$ with $x_k^*$. Finally, pick $y_k\in E$ with $y_k+F=x_k$. Then the map $$T:E\rightarrow E; x\mapsto \sum_k \mu_k(x) y_k$$ is a projection of $E$ onto the span of the $y_k$, and $IT$ will be a projection onto $F$ (unless I've messed something up, which is possible). 


No: If you consider a non continuous functional from a Banach Space, its Kernel is onecodimensional and dense. For example take $l^2(\mathbb{Z})$ and consider the sequence $e_i$ ($(0,..., 1, 0....)$ where the $1$ is in the $i$th position). Complete this to a base (which exists by Zorn's lemma, and it is uncountable since $l^2$ is a banachspace) and consider the subspace generated by the $e_i$ toghether with the elements of the base except one of them. This gives a dense codimension one subspace. 


This is very related to rpotries construction: take a dense, proper subspace and pick a basis $(v_i)$ of that subspace. Now we can adjoin $(w_j)$ such that $(v_i) \cup (w_j)$ is a basis of the whole space. Now the span of all $v_i,w_j$ but finitly many $w_j$ is a dense subspace of finite codimension. So it can not be closed. 


As already recalled, a kernel of any noncontinuous linear form is a dense hyperplane, and noncontinuous forms exist in infinite dimension as a consequence of the existence of Hamel basis. That said, it's worth recalling a relevant fact in the affirmative direction, which is a corollary of the open mapping theorem:
Btw, the property of being complemented has also a particular characterization for those subspaces that are images of operators: the image of $R:X\to Y$ is complemented if and only if $R$ is a right inverse, meaning that there is a bounded operator $L:Y\to X$ such that $LR=I$. A linear projector onto the subspace it is then $RL$. 


It is equi to Choice axiom. If you do not like the Axiom, then every linear functional is continuous. If Not, then HahnBanach theorem is true. It is up to you. Oleg Reinov 


Another counterexample: Take $C[1,1]$ as the Banach Space. Define $ T(f)=\int_{0}^{1} f  \int_{1}^{0} f $. Clearly $T$ is a (continuous) linear functional whose kernel is a (closes) onecodimensional subspace. But there is no complementary subspace: in fact, the complementary subspace should morally be generated by the function which is identically $1$ on $[0,1]$ and $1$ on $[1,0]$, but this function is not continuous. 

