existence of finite-dimensional subspace containing a sequence

Question

Let $X$ be a normed space and suppose we have a sequence $(x_n)$ which is bounded in $X$. Is it possible to find a finite-dimensional subspace $Y$ of $X$ such that $(x_n)\subset Y$?

Answer 1

No, certainly not. For instance, let $(w_n)$ be any sequence of linearly independent elements of $X$ and let $x_n=w_n/\|w_n\|$. Then $(x_n)$ is bounded by linearly independent, so it cannot be contained in any finite-dimensional subspace.

Answer 2

Let us set $E$ to be the subspace of real sequences in $\mathbb{R}^\mathbb{N}$, together with the norm $||(u_n)_n||=sup_n(|u_n|)$. Consider the vectors of $E$ which consist of the sequences $u^i$ defined by $u^i_j=\delta _{i,j}$. Then all the sequences $u^i$ have norm $1$ but the space they generate is infinite dimensional inside $E$ (it is the subspace of all sequences that are eventually 0).

Edit: I replaced "$\mathbb{R}^\mathbb{N}$" with the space of bounded real sequences.