Subspaces of finite codimension in Banach spaces
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 $I-T$ 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 one-codimensional 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 banach-space) 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.
As already recalled, a kernel of any non-continuous linear form is a dense hyperplane, and non-continuous 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:
A linear subspace in a Banach space, of finite codimension, and which is the image of a Banach space via a linear bounded operator, is closed.
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 $RL$.