Working on a finite subspace Let $X$ be a normed space and $L$ a finite dimensional linear subspace.
I need to show that there exists finitely many $l_1,.....l_n \in L$ and $f_1,.....f_n \in X'$ such that $$ l = \sum_{i=1}^{n} f_i(l)l_i$$ for all $l \in L$
Since L is finite it has as basis, so there are $l_1,.....l_n \in L$ with coefficients $f_1,.....f_n$ depending on $l$ such, that we have $$ l = \sum_{i=1}^{n} f_i(l)l_i$$
But how do I show that those $f_i$ are in $X'$??
More I need to show that there exists a continuous projection $P: X \to L$. 
So I need to show that $P(X)=L$ and $P^2=P$. Can I define the projection the following? $$ Px = \sum_{i=1}^{n} f_i(x)x_i$$
P continuous: $$||\sum_{i=1}^{n}f_i(x)l_i||_L \leq \sum_{i=1}^{n}||f_i(x)l_i||_L \leq \sum_{i=1}^{n}\|f_i\|_{X'}\|x\|_X \|l_i\|_L \leq .....$$
I am very thankful for help?

All that is left is, to show that $P^2 = P$.
Since the $f_i$ are defined on $L$ in such a way that $f_i(x)$ is the unique coefficient of $l_i$ in the representation $x = f_1(x)l_1+...+f_n(x)l_n$ and we have $l_i = 0l_1+...+1l_i+...+0l_n$ we have $f_i(l_j) = 1$ if $j=i$ and $f_i(l_j) = 0$ if $i\neq j$.
This yields $$P^2x = P(\sum_{i=1}^nf_i(x)l_i) = \sum_{j=1}^nf_j\left(\sum_{i=1}^nf_i(x)l_i\right)l_j = \sum_{j=1}^n\sum_{i=1}^nf_i(x)f_j(l_i)l_j = \sum_{j=1}^nf_j(x)l_j = Px,$$
where we used that each summand vanishes unless $i=j$ in the second last equality.
