Subsequences of Basis

If $E$ be a Banach space and $(e_n)$ is a Schauder basis of $E$ then any subsequence $(e_{n_k})$ of $(e_n)$ is a basic sequence i.e. $(e_{n_k})$ is a Schauder basis of $M=\overline{\operatorname{span}\{e_{n_k}:k\in\mathbb N\}}$.

I am trying to prove the proposition above. I know that $(e_{n_k})$ is a Schauder basis of $M$ iff $M=\overline{\operatorname{span} \{ e_{n_k} :k\in\mathbb N\}}$ and there exists $c\geq 1$ such that $\left\|\sum_{k=1}^n\lambda_ke_{n_k}\right\|\leq c\left\|\sum_{k=1}^m\lambda_ke_{n_k}\right\|$ for all $(\lambda_k)\subset \mathbb R$ and $n<m$.

First part is our hypothesis. For the second one can I directly say that we have $\left\|\sum_{k=1}^n\lambda_ke_{n_k}\right\|\leq c\left\|\sum_{k=1}^m\lambda_ke_{n_k}\right\|$ since $\left\|\sum_{i=1}^n\lambda_ie_i\right\|\leq c\left\|\sum_{i=1}^m\lambda_ie_i\right\|$ is true for the supersequence?

I appreciate any fixing. Thanks a lot and merry xmas.

$\left\|\sum_{k=1}^n\lambda_ke_{n_k}\right\|\leq c\left\|\sum_{k=1}^m\lambda_ke_{n_k}\right\|$ follows from $\left\|\sum_{i=1}^n\lambda_ie_i\right\|\leq c\left\|\sum_{i=1}^m\lambda_ie_i\right\|$ by just taking coefficients of $e_n$ with $n \notin \{n_1,n_2,\cdots\}$ equal to $0$.

As an example suppose $(e_{n_k})=\{e_2,e_4,\cdots\}$. Putting $\lambda_n=0$ for all odd values of $n$ the inequality $ \|\sum\limits_{k=1}^{2n}\lambda_ke_k\|\leq \|\sum\limits_{k=1}^{2m}\lambda_ke_k\|$ gives $ \|\sum\limits_{k=1}^{n}\lambda_{2k}e_{2k}\|\leq \|\sum\limits_{k=1}^{m}\lambda_{2k}e_{2k}\|$ which is same as $ \|\sum\limits_{k=1}^{n}\mu_ke_{2k}\|\leq \|\sum\limits_{k=1}^{m}\mu_ke_{2k}\|$ for all choices of $\mu_1,\mu_2,\cdots $