Let us denote $l^p(A)=L^p(A,\mu)$ for $1\leq p\leq \infty$, where $A$ is a discrete set with the counting measure $\mu$. I'm unable to understand how this is true, that $$l^p(D)=\mathbb{R}^d$$ where $D=\{0,1,\cdots,d-1\}$.
Can anybody kindly provide an explanation to this?
Moreover, we have the following proposition:
Show that $l^\infty(\mathbb{N})$ is not separable.
Proof: Note that $l^\infty(\mathbb{N})=\cup_{n\in\mathbb{N}} n^\mathbb{N}$ is the set of bounded sequences. In particular, the continuum $2^\mathbb{N}\subseteq l^\infty(\mathbb{N})$ is discrete subset, indeed for every $x,y\in 2^\mathbb{N}$ distinct, $\lVert x-y\rVert_\infty = \sup_{u} |x(n)-y(n)|=1$.
I'm also working on understanding the proof to this, but I guess I'm already stuck on grasping the definition of $l^p$.
I will answer the two points separately.
For the first point, note that as mentioned by @Mittens, there is a reason why the set of all functions $A\rightarrow B$ is sometimes denoted by $A^B$: if $B$ is a discrete set, then $A^B \cong A^{\vert B\vert}$ in the sense of a bijections (i.e. relabelings). This works as follows: if $B = \{b_1, \dots, b_n\}$, then we represent a function $f\in A^B$ by the vector $(f(b_1), \dots, f(b_n))\in A^n$. The other way around naturally works as well.
This answers that $\mathbb R^D$ can be viewed as $\mathbb R^d$. But this does not complete the answer, because $L^p(D, \mu)$ only looks as measurable functions that have a $p$-th integrable power. However, since $D$ is finite, the $\sigma$-algebra on it is simply the power set, so that every function is function. Furthermore, one has $$ \int \vert f(x)\vert^p \mu(dx) = \sum_{i=0}^{d-1} \vert f(i)\vert^p \leq d\cdot \sup_{i=0,\dots, d-1} \vert f(i)\vert^p < +\infty. $$ This proves that every $f\in \mathbb R^d$ is measurable and has a $p$-th integrable power, so that indeed $$ \ell^p(D) = \mathbb R^d. $$
Now concerning your second question. To be honest, I do not know what you mean by $n^\mathbb N$. Instead, I choose to prove the non-separability by a similar argument, but without using that notation. First, note that $$ \ell^\infty(\mathbb N) = \{ (x_n)_{n\in\mathbb N}\in \mathbb R^\mathbb N\;:\; \sup_n \vert x_n\vert < +\infty\}. $$ In particular, $$ \{0, 1\}^\mathbb N \subseteq \ell^\infty(\mathbb N). $$ Furthermore, note that for any two $x,y\in \{0,1\}^\mathbb N$ such that $x\neq y$, there is at least one entry that differs, so that $$ \Vert y - x\Vert_\infty = \sup_n \vert y_n - x_n\vert \geq 1. $$ (We even have $=1$, but that is not relevant.) Now, since $\vert \{0,1\}^\mathbb N\vert = 2^\mathbb N$, we conclude that there are uncountably many elements in $\ell^\infty(\mathbb N)$ that are at distance at least one from each other. In particular, $\ell^\infty(\mathbb N)$ is not separable.