Could someone please prove these equivalences about discrete simplicial sets?
(1) A discrete simplicial set is a simplicial set with no nondegenerate simplices except in degree 0.
(2) Equivalently, it is a simplicial set in which all the face and degeneracy maps are isomorphisms.
(3) Equivalently it is a simplicial set which is isomorphic to a constant functor $\Delta^{op}\to \mathtt{Set}$.
2 clearly implies 1: every $x\in X_n$ for $n>0$ is in the image of any degeneracy map $X_{n-1}\to X_{n}$ since the latter is an isomorphism by assumption; hence $x$ is degenerate (by definition).
Let us show that 1 implies 2: let $X\colon \Delta^{op}\to \mathrm{Set}$ be a simplicial object with no nondegenerate cells except in degree $0$. Given any $n$-simplex $x\in X_n$, it will be in the image of a degeneracy map $X_{n-1}\to X_n$; thus by induction (by composing degeneracies) that means that the unique (composition of) degeneracy map $t_n\colon X_0\to X_n$ is surjective. It is also injective because any (composition of) face map $X_n\to X_0$ is a retraction of $t_n$. Hence $t_n$ is a bijection, i.e. an isomorphism. Now observe that for any map $f\colon [m]\to [n]$ in $\Delta$ we have $X_ft_{n}=t_m$, hence $X_f$ is also an isomorphism. Hence $X$ maps all maps in $\Delta$ to isomorphisms; in particular all face and degeneracy maps are isomorphisms.
The equivalence between 2 and 3 is a general fact about functors:
If $F$ is isomorphic to a constant functor $G$ via an isomorphism $h\colon F\xrightarrow{\cong} G$ then for every map $f\colon a\to b$ we have $h_bF(f)= G(f)h_a=h_a$. Since $h_a$ and $h_b$ are isomorphisms, so is $F(f)$.
If further we assume that the category $A$ has an initial object $I$, then the converse holds: a functor $f\colon A\to B$ (to any category $B$) is isomorphic to a constant functor if it maps all maps to isomorphisms. To see this, take $G$ to be the constant functor with value $F(I)$. There is a canonical natural transformation $j\colon G\to F$ given on $a\in A$ by $F(i_a)\colon G(a)=F(I)\to F(a)$, where $i_a\colon I\to a$ is the unique map. If $F$ maps all maps to isomorphisms, then in particular each component $j_a = F(i_a)$ is an isomorphism, hence $j\colon G\to F$ is a natural isomorphism.
To apply this to $A=\Delta^{op}$, $B=\mathrm{Set}$, observe that $[0]$ is an initial object in $\Delta^{op}$.