Relative compactness criterion in a space with a basis 
Let $X$ be a Banach space with basis. Prove that $M$ is relatively compact if and only if $M$ is bounded and for any $\varepsilon> 0$ there exists $N$ such that $\|R_N (x)\|<\varepsilon$, for any $x$ from $M$ and $R_N (x)$ is the remainder in the expansion of $x$ in the basis.

Sufficiency: we introduce a finite-dimensional subspace $L = \text{span} \{e_k: k = 1, ..., N\}$, then each $x\in M$ can be associated with an element $x_N\in L$, and $\|x − x_N\|< \varepsilon$ by assumption. The set $L$ is bounded, and therefore it is relatively compact and there exists a finite $\varepsilon$-network for it. But then it will be a finite $2\varepsilon$-network for $M$.
My question is about proof of necessity. I can’t do it. From the existence of a finite $\varepsilon$-network $y_j$ we conclude that $M$ is bounded, and also that for some $j$ $\|x − y_j\| <\varepsilon$. In addition, there is a number $N$ such that for all $y_j$ $\left\|\sum\limits_{k=N}^\infty c_k^je_k\right\| <\varepsilon$. Then for any $x$ $$\left\|\sum\limits_{k=N}^\infty c_ke_k\right\|\leq\left\|\sum\limits_{k=N}^\infty (c_k − c_k^j) e_k\right\| + \left\|\sum\limits_{k=N}^\infty c_k^je_k\right\|.$$
And the first norm on the right-hand side is not evaluated in terms of $\|x − y_j\|$. Can you please suggest how to prove this part. Or maybe it is generally incorrect and there is a counterexample?
 A: First, a very minor correction to your argument for sufficiency. $L$ itself isn't bounded, however this is a non-issue since $L \cap M$ is bounded and so working with this set instead makes your argument work.
Now for necessity. This argument is essentially a complete version of the one you've started but I present the whole thing for clarity.
Let $\{e_i: i \in \mathbb{N}\}$ be a Schauder basis for $X$. Let $P_N$ be the projection onto $\{e_1, \dots, e_{N-1}\}$. The key point is that it is a consequence of the open mapping theorem that $C = \sup_N \|P_N\| < \infty$. Fix $\varepsilon > 0$ and let $\alpha = \frac{\varepsilon}{2+2C}$. 
By relative compactness, pick an $\alpha$-net for $M$ $\{y_1, \dots, y_k\}$. Then for each $n$ we have that $y_n = \sum_{i=1}^\infty c_i^n e_i$ for some constants $c_i^n$. In particular, there is an $N$ such that $$\max_{j = 1, \dots, k} \bigg\|\sum_{m=N}^\infty c_i^n e_i\bigg\| < \frac{\varepsilon}{2}$$ 
We also have that there is a $y_j$ such that $\|x- y_j \| < \alpha$. Fix a choice of $j$ satisfying this bound. We then have
$$\|R_N(x)\| = \|R_N(x-y_j) + R_N(y_j)\| \leq \|R_N(x-y_j)\| + \|R_N(y_j)\| < \|R_N(x-y_j)\| +  \frac{\varepsilon}{2}$$
From here, it is straightforward to finish since $R_N = \operatorname{Id} - P_N$. Hence $$\|R_N\| \leq 1 + \|P_N\| \leq 1+ C$$
and so $$\|R_N(x)\| \leq \|R_N(x - y_j)\| + \frac{\varepsilon}{2} < (1+C) \alpha +  \frac{\varepsilon}{2} = \varepsilon$$
as desired. 
