Banach Space Sequence Equivalent to standard basis of $\ell_1$ Definition
We define a sequence $(y_n)_n$ in $S_X$ (the unit circle) to be equivalent to the standard basis of $\ell_1$ if there exist $C,c>0$ such that for every $k\in\mathbb{N}$ and $a_1,...,a_k\in\mathbb{R}$ we have $c\sum_{i=1}^k|a_k|\leq{}\|\sum_{i=1}^ka_ky_k\|\leq{}C\sum_{i=1}^k|a_k|$.
Question
Suppose $x_n$ is a sequence in $S_X$ that is linearly independent as a set and $x\overset{\ast}{\rightharpoonup}0$ in the weak topology. 
I need to prove that if $(y_n)_n\in{}S_X$is equivalent to the standard basis of $\ell_1$ then $\liminf\limits_{n\rightarrow\infty}\|x_n-y_n\|>0$.
Some Remarks
I know that if I had $y_n\overset{\ast}{\rightharpoonup}y$ then the statement reduces to showing 
$\|y\|>0$ since $\liminf\limits_{n\rightarrow\infty}\|x_n-y_n\|\geq{}\|x-y\|$ but even then $\ell_1$ can be embedded isomorphically in $span(y_n)$ and weak convergence guarantees $\|y\|=0$.
 A: Attempt at answer from @DavidMitra 's hint
If $\liminf\limits_{n\rightarrow\infty}\|x_n-y_n\|=0$ then there exists a subsequence such that $\lim\limits_{i\rightarrow\infty}\|x_{n_i}-y_{n_i}\|=0$. For an arbitary $\epsilon>0$ we can pick a subsequence $n_1,...,n_k$ such that $\|x_{n_i}-y_{n_i}\|<\epsilon$ for every $i$ in $1,...,k$. 
Then $\|\sum_{i=1}^ka_ix_{n_i}\|=\|\sum_{i=1}^ka_i(x_{n_i}-y_{n_i})+\sum_{i=1}^ka_iy_{n_i}\|\leq{}\|\sum_{i=1}^ka_i(x_{n_i}-y_{n_i})\|+\|\sum_{i=1}^ka_iy_{n_i}\|\leq{}\epsilon\sum_{i=1}^k|a_i|+C\sum_{i=1}^k|a_i|=(\epsilon+C)\sum_{i=1}^k|a_i|$
For the other part: $c\sum_{i=1}^k|a_i|\leq{}\|\sum_{i=1}^ka_iy_{n_i}\|=\|\sum_{i=1}^ka_i(x_{n_i}-y_{n_i})+\sum_{i=1}^ka_ix_{n_i}\|\leq{}\|\sum_{i=1}^ka_i(x_{n_i}-y_{n_i})\|+\|\sum_{i=1}^ka_ix_{n_i}\|\leq\epsilon\sum_{i=1}^k|a_i|+\|\sum_{i=1}^ka_ix_{n_i}\|$ giving $(c-\epsilon)\sum_{i=1}^k|a_i|\leq{}\|\sum_{i=1}^ka_ix_{n_i}\|$.
Choosing $\epsilon<c$ gives us a standard basis of $\ell_1$ that is weakly null which is a contradiction.
