Let $e^{(k)}=(0, \ldots, 0, 1, 0, \ldots)$ where the $k$th entry is $1$ and the rest are $0$s. Show that $\{e^{(k)}: k\ge 1\}$ is closed as a subset of $l_1$.
My attempt:
A theorem in my textbook states that given a set $F$ in $(M, d)$, the following are equivalent:
(i) $F$ is closed.
(ii) If a sequence $(x_n)\subset F$ converges to some point $x\in M$, then $x\in F$.
Let $(x_n)$ be any sequence of points (sequences) in $\{e^{(k)}: k\ge 1\}$. $d(x_m, x_n)=2$ if $x_m\ne x_n$ and $d(x_m, x_n)=0$ if $x_m=x_n$. So, $(x_n)$ is Cauchy iff it is eventually equal to some element of $\{e^{(k)}: k\ge 1\}$. Thus, if $(x_n)\subset\{e^{(k)}: k\ge 1\}$ converges to some point $x\in l_1$, then $x\in\{e^{(k)}: k\ge 1\}$. So, (ii) is true, thereby (i) is true and $\{e^{(k)}: k\ge 1\}$ is closed.
Is this correct? I'm very new to analysis, so I think I may have missed something.
You have not missed anything. As a subset of $l^1$, the $E = \{e^{(k)}\}$ form an isolated set, since around every point $e^k$ you can find $\epsilon > 0$ such that $0 < d(x,e^{k}) < \epsilon \implies x \notin E$.
It is not difficult to see, from a similar argument, why every convergent sequence in an isolated set must be eventually constant, and therefore every isolated set is closed in a metric space. You have just applied this corollary to the given set.