Here is the whole comment exchange under this answer (currently deleted by the OP.). (At least the part that I can see. Moderators would be able to tell you whether there were additional comments, which were deleted.)
I am afraid that you loose the generality when restricting $y_i^\epsilon\in L$ so that the function $F$ is well defined. Nevertheless, I like your solution. -- Svetoslav
@Svetoslav: couldn't we remove the loss of generality but otherwise keep the proof the same by using the n-fold wedge map, instead of the determinant? -- ziggurism
@ziggurism I am not familiar with this map and its properties. -- Svetoslav
I think it provides a concise elegant proof of OP's question, so you may be interested in learning more. http://en.wikipedia.org/wiki/Wedge_product -- ziggurism
You are losing generality by assuming that $(y_i^\epsilon)_{i\in\mathbb N}\subset L$. In such a way you are answering the OP question at most in the case when $X$ is finite dimensional. -- Svetoslav
@Svetoslav in the very begining I'm saying that it is enough to answer the question for the linear span of vectors x_1, ...,x_n, not the whole X. -- Norbert
The problem is that the $y_i^\epsilon$ are not supposed to be also in the span of $x_1,..,x_n$. Then you cannot represent $y_1^\epsilon,...,y_n^\epsilon$ with the same functionals $c_1,...,c_n$ for every $\epsilon$ in the sequence that you construct. Or you have to justify how you would deal with this in the general case. -- Svetoslav
The problem asks to find one tuple $y_i$. While it doesn't require that the tuple be in the span of the $x_i$, it doesn't require that they not be either, so finding any independent tuple solves the problem, even if it is in the span. There could also be tuples not in the span, this is ok. I think there is no loss of generality. -- ziggurism
@ziggurism What do you mean by "..asks to find one tuple $(y_i)_{i\in\mathbb N_n}$" ? The problem asks to find one $\epsilon>0$ s.t for $\textbf{all}$ tuples $(y_i)_{i\in\mathbb N_n}$ with $\|y_i\|\leq \epsilon$ the vectors $x_1+y_1,..,x_n+y_n$ are linearly independent. This translates to find one $\epsilon>0$ s.t $x_1+B_\epsilon,...,x_n+B_\epsilon$are linearly independent (informally said as the above are sets). And the ball $B_\epsilon$ is in the possible infinite dimensional normed space $X$ and thus not compact, as it is if you consider the finite dimensional $L$. -- Svetoslav
Norbert, your contradiction argument also fits to "there exists $\epsilon>0$ s.t for all tuples $(y_i)_{i\in\mathbb N_n}$ with $\|y_i\|\leq \epsilon,\,i=1,..,n$ ", so you also prove that $x_1+B_\epsilon,...,x_n+B_\epsilon$ are "linearly independent" for some small $\epsilon$. I'm not trolling here, just want to understand what is going on. -- Svetoslav