Let $(X, d_{l^1})$ be the metric space from Exercise 1.1.15. For each natural number $n$, let $e^{(n)} = (e^{(n)}_j)_{j=0}^\infty$ be the sequence in $X$ such that $e^{(n)}_j : = 1$ when $n=j$ and $e^{(n)}_j:=0$ when $n \not= j$. Show that the set $\{e^{(n)} : n \in \mathbb{N}\}$ is a closed and bounded subset of $X$, but is not compact.
The set is bounded since $d_{l^1}(e^{(j)}, e^{(k)}) \le 2$ for all $j,k \in \mathbb{N}$ (i.e., there exists a ball which contain the set).
Consider the sequence $(e^{(n)})_{n=0}^\infty$. This sequence is not Cauchy because $d_{l^1}(e^{(j)},e^{(k)}) = 2$ for every $j \not= k$. Also, there exists no subsequence which is convergent in this set. Thus, it is not compact.
How can I show closedness? Is my proof correct?
What you have done so far is correct. To show that the set is closed suppose $x$ belong to its closure. Then $(e^{(n_k)})$ converges to $x$ for some $(n_k)$. If $n_k$'s are distinct then $2=d(e^{(n_k)}, e^{(n_j)})$ must tend to $0$ as $j,k \to \infty$ which is a contradiction. Thus there is an eventually constant sequence converging to $x$ and Hence $x$ belongs to the set.