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For instance, let $M = \{x_i: x_i \to 2\}$ be a subset of $l^\infty$ with the infinity metric. Is M closed in $l^\infty$?

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    Answer to question in title: no, just consider open unit ball in $\mathbb{R}^n$. For your particular $M$, suppose $x^m$ is a sequence in $M$ converging to $x$. Completeness implies $x\in \ell^{\infty}$. Now suppose that $x_n \not\to 2$. Then there is $\epsilon>0$ such that there is a subsequence $x_{n_k}$ with $|x_{n_k}-2|>\epsilon$ for all $k$. Do you see why this is a contradiction? – 2017-02-07
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    $M$ is not bounded. For each $n\in \Bbb N$ consider the sequence $\{x_{n,m}\}_{m=1}^\infty$ defined by $$x_{n,m}=\begin{cases}n\text{ if }m=1\\2\text{ otherwise}\end{cases}$$ Then $\|x_n\|_\infty=n$. – 2017-02-07

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Let $\ell_\infty$ denote the space of all real and bounded sequences, normed by :

$$\forall x\in\ell_\infty,\,\Vert x\Vert_\infty=\sup\{\vert x_n\vert\;\,n\in\mathbb{N}\}$$

Let $c$ the subspace of convergent sequences. It is well known that $c$ is a closed subspace of $\ell_\infty$.

The linear map $c\to\mathbb{R},x\mapsto\lim x$ is continuous, because :

$$\forall x\in c,\vert\lim x\vert\le\Vert x\Vert_\infty$$

Hence, if $B$ denotes the set of all sequences converging to $2$, then $B$ is closed in c because it is the pre-image of $\{2\}$ (which is a closed subset in $\mathbb{R}$), and thus closed in $\ell_\infty$

However, as other comments suggest, $B$ is not a bounded subset of $\ell_\infty$.