$\| \sum_{k=1}^{m}a_ke_k \| \leq K \| \sum_{k=1}^n a_k e_k \|$ for $(a_k)$ and $m,n$ such that $m \leq n$ implies $(e_k)$ is linearly independent I am reading 'Topics in Banach Space Theory'. 

A sequence $(e_k)_{k=1}^{\infty}$ is non-zero elements of a Banach space $X$ is basic if and only if there is a positive constant $K$ such that 
  $$\| \sum_{k=1}^{m}a_ke_k \| \leq K \| \sum_{k=1}^n a_k e_k \|$$
  for every sequence of scalars $(a_k)$ and all integers $m,n$ such that $m \leq n$.

We want to prove the direction $(\Leftarrow)$.
Proof: Denote $E$ as the linear span of $(e_k)_{k=1}^{\infty}$. The inequality implies that the vectors $(e_k)_{k=1}^{\infty}$ are linearly independent. 
Question: How to prove the bolded sentence? From what I know, to show a linearly independence of a set containing infinitely many elements, we need to show that for any finite index set $I$, $\sum_{i \in I}a_i e_i = 0 \Longrightarrow a_i = 0$ for all $i \in I.$ I do not know how to apply the inequality to reach the conclusion. 
UPDATE: I have an idea, but would like someone to verify it: 
Suppose $I$ is a finite set, say $|I| = n$, and $(a_i)$ is a finite sequence of scalars such that $\sum_{i \in I}a_i e_i = 0.$ Since $I$ has $n$ elements, its elements can be ordered, say, $i_1 \leq i_2 \leq i_3 \leq ... \leq i_n$. Let $n = i_n$ and $m = i_{n-1}$ in our assumption. But $K$ does not allow me to have $a_{i_n}e_{i_n}=0.$
 A: Suppose you have a sequence $(a_k)$ of scalars such that


*

*there is an $n\in \mathbb{N}$ such that $k > n \implies a_k = 0$, and

*$\displaystyle \sum_{k = 1}^{\infty} a_k e_k = 0$.


Fix an $n$ according to the first point. If $n = 0$, you're done. Otherwise, choose $m = 1$ first. Then
$$\lVert a_1 e_1\rVert \leqslant K\Biggl\lVert \sum_{k = 1}^n a_k e_k\Biggr\rVert = K\lVert 0\rVert = 0$$
implies $a_1e_1 = 0$, and since $e_1\neq 0$ it follows that $a_1 = 0$. If $n = 1$, you're done, otherwise consider $m = 2,\,\dotsc,\, m = n$ to reach the conclusion $a_k = 0$ for $1 \leqslant k \leqslant n$. Hence we have

If $(a_k)$ is a sequence of scalars with only finitely many nonzero terms, and $$\sum_{k = 1}^{\infty} a_k e_k = 0,$$ then $a_k = 0$ for all $k$.

And that is just the definition of linear independence of the $e_k$.
This is basically the same as your idea, but for your idea, you need to extend the set of scalars (by zeros) to those $k < i_n$ that are not equal to any of the $i_r$ to apply the hypothesis, and you also need to apply the hypothesis multiple times - for each coefficient that you don't a priori know is $0$. If you apply the hypothesis to $m = i_{n-1}$ first, you cannot directly deduce that $a_{i_r} = 0$ for $1 \leqslant r \leqslant n$, only that $a_{i_n} = 0$ and
$$\sum_{r = 1}^{n-1} a_{i_r} e_{i_r} = 0.$$
