Proving closure of set My attempt:
$\overline{B} \subseteq \ell^\infty$ follows from the fact that $B \subseteq \ell^\infty$ and $\ell^\infty$ is closed.
For $\ell^\infty \subseteq  \overline{B}$, we must show every sequence in $\ell^\infty$ can be represented as a limit point of a sequence in $B$. Thus, consider $x = (x_1, x_2, ...) \in \ell^\infty$. If $x$ has a constant subsequence, we are done, so consider when $x$ does not. Let $d_i = (x_1, x_2, ..., x_i, \sup\{x_{i+1}, x_{i+2}, ...\}, \sup\{x_{i+1}, x_{i+2}, ...\}, ...\}$ . Thus, $d_i$ is an element of $B$ as it has a constant subsequence.
Now, consider the case where $x$ does not reach its supremum on any of its elements. Thus, $x$ must have an infinitely increasing or decreasing tail whose elements converge to the supremum. Thus, by nature of the supremum, we can find a $N \in \mathbb{N}$ such that when $n > N$, $||x - d_n|| < \epsilon$. Thus, we need only consider when $x$ has no infinitely decreasing/increasing tails, so reaches its supremum on all tails. Formally, $||x||_\infty$ is reached by some $x_i$ on each tail of $x$. Thus, since $x$ has no constant subsequence, $||x||_\infty$ can only be repeated a finite number of times in the sequence. Thus, we can form a decreasing chain of maximum elements in $||x||_\infty$. Thus, $||x - d_i||$ form an eventually decreasing sequence as $i \rightarrow \infty$. However, at this point I am having trouble proving that $||x - d_i||$ becomes $\epsilon-$close to each other and can only really state that they form a decreasing chain.
For the second question, i.e. showing what $int(B)$ is, I believe $int(B) = \emptyset$. Suppose for a contradiction $B_r(x) \subseteq B$ for some $x \in B, r > 0$. However, consider the sequence $y$ which replaces each constant subsequence $x_{n_k}$ of $x$ with the subsequence $x_{n_k} + r/2^k$ and is the same otherwise. Thus, $||x - y|| = r/2$ so $y \in B_r(x)$ but $y \not \in B$, so this open ball cannot be contained in $B$. Contradiction.
At this point, I would really appreciate any feedback on whether or not I am on the right track for both of these proofs (and if the latter looks complete or not). Thank you!
 A: $\overline{B}=\ell^\infty$.
Proof: Let $x=(x_1,x_2,\ldots)\in\ell^\infty$ and let $\epsilon>0$. Then the set of values $\{x_n:n\in\mathbb{N}\}$ is bounded by $\|x\|_\infty$, hence has a convergent subsequence. Pick any of these limit points, but let's pick $\alpha=\limsup x$ as OP did. Then by definition there is a sequence of indices $n_i$ such that $$i>N\implies|x_{n_i}-\alpha|<\epsilon$$
Construct the sequence $y:=(y_n)$ such that $y_n=\begin{cases}\alpha&\exists i, n=n_i,\\x_n&\forall i, n\ne n_i\end{cases}$. Since $n=n_i$ for an infinite number of times, $y$ has a constant subsequence and so $y\in B$. Also $$\|x-y\|_\infty=\sup_n|x_n-y_n|=\sup_i|x_{n_i}-\alpha|<\epsilon$$ so $y\in B_\epsilon(x)$, and $B$ is dense in $\ell^\infty$.
The interior of $B$ is empty since any sequence $y$ with a constant sub-sequence can be modified by small amounts that destroy the constant sub-sequence. For example, if $y$ has a single constant sub-sequence, then $y+(1/n)_{n>N}$ is within $1/N$ of $y$ but has no constant sub-sequence.
