Proof Mark ${\mathbf{x}}_n=(x_1^{(n)},...,x_m^{(n)})\in {\mathbb{R}}^m$. $|x_k^{(n)}|\le \parallel {\mathbf{x}}_n{\parallel }_{\mathrm{\infty }}\le M$, so $\{{\mathbf{x}}_n\}$ is bounded $\iff \{x_k^{(n)}\}$ is bounded for any $k$. Also $\parallel {\mathbf{x}}_n{\parallel }_{\mathrm{\infty }}\le \sum _{i=1}^m|x_i^{(n)}|$, $|x_k^{(n)}-x_k^{(0)}|\le \parallel {\mathbf{x}}_n-{\mathbf{x}}_0{\parallel }_{\mathrm{\infty }}\le \sum _{i=1}^m|x_i^{(n)}-x_i^{(0)}|$. So $\parallel {\mathbf{x}}_n-{\mathbf{x}}_0{\parallel }_{\mathrm{\infty }}$ could be constrained by $m$ one-dimension sequences. $◻$

Proof Right-handed side. $\mathbf{x}=\sum _{i=1}^m{x}_i{\mathbf{e}}_i$, so $$\begin{array}{rl}\parallel \mathbf{x}\parallel & \le |x_1|\parallel {\mathbf{e}}_1\parallel +...+|x_m|\parallel {\mathbf{e}}_m\parallel \\ & \le \parallel {\mathbf{x}}_n{\parallel }_{\mathrm{\infty }}(\parallel {\mathbf{e}}_1\parallel +...+\parallel {\mathbf{e}}_m\parallel +1)\\ & =C\parallel {\mathbf{x}}_n{\parallel }_{\mathrm{\infty }}\end{array}$$

Left-handed side is proved by contradiction. Assuming that for any $C>1$, there exists $\mathbf{x}\in {\mathbb{R}}^m$ such that $\frac{1}{C}\parallel \mathbf{x}{\parallel }_{\mathrm{\infty }}>\parallel \mathbf{x}\parallel$. For any $n\in {\mathbb{N}}^{+}$, there exists ${\mathbf{x}}_n$ such that $\parallel \mathbf{x}{\parallel }_{\mathrm{\infty }}>n\parallel \mathbf{x}\parallel \ge 0$. Let ${\mathbf{y}}_n=\frac{{\mathbf{x}}_n}{\parallel {\mathbf{x}}_n{\parallel }_{\mathrm{\infty }}}$, then for any $n$, $\parallel {\mathbf{y}}_n{\parallel }_{\mathrm{\infty }}=1$ and $\parallel {\mathbf{y}}_n\parallel =\frac{\parallel {\mathbf{x}}_n\parallel }{\parallel {\mathbf{x}}_n{\parallel }_{\mathrm{\infty }}}<\frac{1}{n}$. $\parallel {\mathbf{y}}_n{\parallel }_{\mathrm{\infty }}=1$ means $\{{\mathbf{y}}_n\}$ is bounded under $\parallel \cdot \parallel$, so it has convergent subsequence $\{{\mathbf{y}}_{n_k}\}$. When $k\to +\mathrm{\infty }$, $\parallel {\mathbf{y}}_{n_k}-{\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}\to 0$. $$\begin{array}{rl}\parallel {\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}\le \parallel {\mathbf{y}}_{n_k}-{\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}+\parallel {\mathbf{y}}_{n_k}{\parallel }_{\mathrm{\infty }}& \to \parallel {\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}\le 1\\ 1=\parallel {\mathbf{y}}_{n_k}{\parallel }_{\mathrm{\infty }}\le \parallel {\mathbf{y}}_{n_k}-{\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}+\parallel {\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}& \to \parallel {\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}\ge 1\\ \parallel {\mathbf{y}}_0\parallel \le \parallel {\mathbf{y}}_{n_k}-{\mathbf{y}}_0\parallel +\parallel {\mathbf{y}}_{n_k}\parallel \le C\parallel {\mathbf{y}}_{n_k}-{\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}+\frac{1}{n_k}& \to \parallel {\mathbf{y}}_0\parallel =0\to \parallel {\mathbf{y}}_0{\parallel }_{\mathrm{\infty }}=0\end{array}$$

They are contradicted. $◻$

Proof Sufficiency. Assuming $E$ is bounded closed, then for any $\{{\mathbf{x}}_n\}\subseteq$ is bounded, so it has convergent subsequence $\{{\mathbf{x}}_{n_k}\}$ and $\underset{k\to +\mathrm{\infty }}{lim}{\mathbf{x}}_{n_k}={\mathbf{x}}_0\in E$.

Necessity. Prove it is bounded first by contradiction. Assuming $E$ is unbounded, meaning for any $n\ge 1$, there exists ${\mathbf{x}}_n\in E$ such that $\parallel {\mathbf{x}}_n\parallel >n$. According to the precondition, $\{{\mathbf{x}}_n\}$ has a convergent subsequence $\{{\mathbf{x}}_{n_k}\}$ and $\parallel {\mathbf{x}}_{n_k}\parallel >n_k\ge k$. So $\{{\mathbf{x}}_{n_k}\}$ is unbounded, contradicted with its convergence. So the assumption is false, $E$ is bounded.

Prove it is closed then. The convergent subsequence $\{{\mathbf{x}}_{n_k}\}$ of $\{{\mathbf{x}}_n\}$ subjects to $\underset{k\to +\mathrm{\infty }}{lim}{\mathbf{x}}_{n_k}={\mathbf{x}}_0\in E$. So if $\{{\mathbf{x}}_n\}$ is convergent, $\underset{n\to +\mathrm{\infty }}{lim}{\mathbf{x}}_n={\mathbf{x}}_0\in E$. So $E$ is closed. $◻$