Write $\Delta[n]$ for the (semi)simplicial object represented by the finite ordinal $[n]=0<1<\cdots <n$.
Here's an argument for proving the (semi)simplicial cone $\mathrm C\Delta[n]$ is isomorphic to $\Delta[n+1]$.
By Yoneda, any arrow $\Delta[n+1]\Rightarrow \mathrm C\Delta[n]$ uniquely corresponds to an element of $(\mathrm C\Delta[n])_{n+1}\cong \bf 1$. Hence there's only one such arrow $\eta:\Delta[n+1]\Rightarrow \mathrm C\Delta[n]$, given by the formula in the proof of the Yoneda lemma. We aim to prove $\eta$ is an isomorphism so it suffices to do this component-wise. Thus we consider $$\eta_{[k]}:\mathrm (\Delta[n+1])_k\longrightarrow (\mathrm C\Delta[n])_k=(\Delta[n])_k\amalg (\Delta[n])_{k-1}.$$ If we partition the domain into arrow with $k\mapsto n+1$ and $k\not\mapsto n+1$ then there's an obvious coproduct of canonical bijections. However it seems unpleasant to prove $\eta_[k]$ actually is this coproduct.
Apart from calculation, is there a slick way to prove this fact about $\eta_{[k]}$? Alternatively, is there a slick way to prove $\mathrm C\Delta[n]\cong\Delta[n+1]$?
Added. The definition from which I must start is $$(\mathrm{C}X)_0=X_0\amalg \mathbf{1},(\mathrm{C}X)_n=X_n\amalg X_{n-1}$$ with the following face maps. $$d_i^{(\mathrm CX)_{n}}=\begin{cases} (X_n\amalg X_{n-1}\overset{d_i^{X_n}\amalg d_i^{X_{n-1}}}{\longrightarrow} X_{n-1}\amalg X_{n-2}) & 0\leq i\leq n-1 \\ (X_n\amalg X_{n-1}\overset{(d_i^{X_n}, 1_{X_{n-1}})}{\longrightarrow} X_{n-1}\amalg X_{n-2}) & i=n \end{cases}$$
Maybe a possible way to solve your problem is to use what a call universal-property-yoneda style.
It is know that a natural transformation $$\eta \colon \mathcal C[-,c] \to F$$ is an isomorphism if and only if the element that correspond to it via the Yoneda-bijection (i.e. $\eta_c(1_c)=x^*$) is an universal element for the functor $F$.
In the case at hand we can use this fact to reduce proving that $\eta$ is an isomorphism to proving that $x^* \in C(\Delta[n])_{n+1}$ is a universal element for the functor $C(\Delta[n])$.
It is not really hard to prove that this element is universal. Let $x \in C(\Delta[n])_k$ be a $k$-simplex (where clearly $k \leq n$). We want to show that there is a unique $\alpha \in \Delta[k,n+1]$ such that $C(\Delta[n])_\alpha(x^*)=x$.
At this point by calculations, using the definition of the structure maps of the functor $C(\Delta[n])$, it should be possible to prove this property, hence that $x^*$ is universal and so that $\eta$ is indeed an isomorphism.