E Banach space $(x_n)\subset E$ such that $ \| x_n \| = 1 $ and $x_n$ weak convergent to $0$ Suppose that $ M \subset E$ finite dimensional subspace. Let E Banach space and $ (x_n)\subset E$  such that $ \| x_n \| = 1 $ and $x_n$ weak convergent to $0$. Suppose that $ M \subset E$ finite dimensional subspace, prove that for each $0 < \delta < 1$ there exists $ n_0 \in \mathbb{N}$ such that $ \forall x \in M$ , $\lambda \in F \; \; (\mathbb{R} \;
\; \textrm{or}\;\; \mathbb{C}) $ and  for all $n \geqslant n_0$,
$$  \| x + \lambda x_n \| \geqslant (1-\delta) \|x_n\| $$
My proof is: $ M \subset E$ finite dimensional so $S_M$ (ball surface) compact then $ S_M$ is totally bounded hence,
$$ S_M \subset \cup  _{j=1} ^{p} B(a_j,\delta)$$
where $ a_1 ,a_2 , \cdots , a_p \in S_M$. By Hahn Banach thereom, we have for all $ 1 \leqslant j \leqslant p \; $, $\; \phi _{j} (a_j) = 1$ where $ \phi _1 , \phi , _2 \cdots , \phi _n \in S_{E^{'}}$. Also $ x_n \rightarrow 0$ weak convergent then we have for all $ j \in \{ 1, \cdots ,p\} $
$$ \phi _j (x_n) \rightarrow 0 \textrm{weak convergent} $$
and then
\begin{equation}
\begin{split}
\| x + \lambda x_n \| & = \| x + a_j - a_j + \lambda x_n \|\\
&\geqslant | \| a_j + \lambda x_n \| - \| a_j - x\| |\\
& \geqslant | \phi _j (a_j + \lambda x_n) - \delta | \\
& = | \phi _j (a_j) + \lambda \phi_j (x_n) - \delta | \\
& =| 1 + \lambda\phi_j(x_n) -\delta|
\end{split}
\end{equation}
i stack here. Could you any hint ?
 A: *

*I assume the following "uniformity" is desired: For any $\lambda\in \Bbb K$ and $\delta\in(0,1)$ there is a $n(\delta;\lambda)$ so that for all $x\in M$ one has $$\|x+\lambda x_j\| ≥ \|x\|(1-\delta)$$
whenever $j≥n$. This change is necessary because the statement clearly fails if the $n$ should exist for all $\lambda$ simultaneously (eg $E=M=\Bbb R$, $x_j=-\frac1j$) you have for $\lambda = n(\delta)$ that the statement fails for $x=1$).

*Suppose the statement holds for all $x$ with $\|x\|=1$ and all $\lambda$. Then it holds for all $x$ and all $\lambda$ since
$$\|x+\lambda x_j\| =\|x\|\cdot\|\frac{x}{\|x\|}+\frac{\lambda}{\|x\|}x_j\| ≥\|x\|\cdot \left(\|\frac{x}{\|x\|}\|(1-\delta)\right)=\|x\|(1-\delta)$$
So its enough to show the existence of this $n$ for all $x$ with $\|x\|=1$ and $\lambda$ arbitrary but fix.
Let $x\in M$, $\lambda\in\Bbb K$ be arbitrary but fixed. By Hahn Banach there is some norm one functional $f$ so that $f(x)=\|x\|$. Then
$$\|x+\lambda x_j\|≥ |f(x+\lambda x_j)| ≥ f(x)-|\lambda |\ |f(x_j)|=\|x\|- |\lambda|\ |f(x_j)|$$
(where it was used that $g(z) ≤ \|g\|\cdot \|z\|$ for any $g,z$) now since $x_j$ converges to $0$ weakly you have that the right-hand side converges to $\|x\|$. Put a $\liminf$ on both sides of the inequality to get that there is an $N(x;\epsilon)$ so that
$$\|x+\lambda x_j\| ≥ \liminf_n \|x+\lambda x_n\| +\epsilon ≥ \|x\|-|\lambda| \epsilon+\epsilon$$
where $\epsilon$ is arbitrary $>0$. Taking $\epsilon = \delta\frac{\|x\|}{1-|\lambda|}$ would recover the statement if there weren't that additional condition that this should work for all $x\in M$ simultaneously.
Now cover the unit sphere of $M$ with finitely many balls of radius $r$ - to be determined later. Suppose $y_1,...,y_k$ are the centres, do the example above for each $y_i$ separately to find that
$$\|y_i-\lambda x_j\| ≥ 1(1-\widetilde\delta)$$
when $j>n$ for $n(\lambda;\widetilde\delta)$ finite. $\widetilde\delta$ is some number in $(0,1)$ that can be chosen freely. Then for any $x$ in the unit sphere there is a $y_i$ so that $\|x-y_i\|<r$ whence
$$\|x-\lambda x_j\| ≥ 1-\widetilde\delta-r$$
for all $x$ in the unit sphere and $j>n(\lambda;\widetilde \delta)$. Now take $r=\delta/2$, $\widetilde \delta = \delta/2$ and you are done.
