Compactness and Schauder basis Let $X$ be a (real) Banach space and $(x_{n})_{n\geq 1}\subset X$ a Schauder basis, that is to say, for each $x\in X$ there are a unique sequence $(a_{n})_{n\geq 1}\subset \mathbb{R}$ such that
$$
x =\sum_{n\geq 1}a_{n}x_{n}.
$$
ADDED (EDITED):  We assume $\|x_{n}\|=1$ for each $n\geq 1$.
In fatc, for each $n\geq 1$, taking a sequence $(x_{n}^{*})_{n\geq 1}\subset X^{*}$ (the topological dual of $X$) such that $x_{m}^{*}(x_{n})=\delta_{mn}$ (The Kronecker symbol) we hace $a_{n}=x_{n}^{*}(x)$.
My question is the following: Assume $K\subset X$ is compact. Can we state that given any $\varepsilon>0$ there is $n_{0}:=n_{0}(\varepsilon)$ such that $|x_{n}^{*}(x)|\leq \varepsilon $, for all $n\geq n_{0}$ and $x\in K$?
I know that the above is true for the $\ell_{p}$ spaces ($1\leq p<\infty$), the usual Banach space of the sequeces such that $\sum_{n\geq 1}|x_{n}|^{p}<\infty$.
Many thanks in advance for your comments.
 A: Assume there is $\varepsilon >0$, a sequence $(y_n) \in K^n$ and a strictly increasing function $\varphi:\mathbb N \to \mathbb N$ such that :
$$\forall n\in\mathbb N , \left|x_{\varphi(n)}^*(y_n)\right|>\varepsilon$$
Since $K$ is compact, we can assume WLOG that $(y_n)$ converges to some $y\in K$. Then, we have:
$$y = \sum_n x^*_n(y)x_n$$
This series converges, we have $\|x_n^*(y)x_n\| = |x_n^*(y)| \to 0$. However:
\begin{align}
\varepsilon & \leq |x_{\varphi(n)}^*(y_n)|\\
&\leq \|x^*_{\varphi(n)}\|\|y_n-y\| + |x^*_{\varphi(n)}(y)|
\end{align}
Since the $x^*_n$ are uniformly bounded, this is a contradiction, as the RHS goes to zero.
Edit We want to prove an assertion of the form : $$\forall \varepsilon >0,\exists n_0\in\mathbb N, \forall n\geq n_0, P(n)$$ By contradiction, we assume the negation of this, ie : $$\exists \varepsilon >0, \forall n_0 \in\mathbb N, \exists n\geq n_0, \neg P(n)$$
This last assertion is equivalent to the existence of a strictly increasing function $\varphi:\mathbb N \to \mathbb N$ such that $\forall n\in\mathbb N, \neg P(\varphi(n))$. I think this form is usually easier to work with.
