I recently came across the following statement, and still can't prove it:
Statement: Suppose $X$ is a separable,closed subspace of $L^1(G)$, where $G$ is a locally compact group. Since $X$ is separable, there is a sequence $(x_n)_{n \in \mathbb{N}}$ which spans a dense subspace of $X$ and $\lim\limits_{n \rightarrow \infty}||x_n||=0$,
My problem is that I feel this statement is false: Suppose such a sequence exists. Then you can find $N$ such that $||x_n||<1$ for all $n>N$. Let $M=\max\{||x_1||,\cdots,||x_N||\}$. Then any element outside the ball of radius $M+1$ can't be "close" to any $x_n$, and hence $x_n$'s can't be dense. Can someone explain to me if my reasoning is correct, or if there is a flaw somewhere. Also, if my reasoning is incorrect and the original statement is true- I would appreciate a hint towards the proof.