As FShrike notes in the comments, you'll need to use the Eilenberg-Zilber Lemma: For any simplex $x$ in $X_n$, there is a unique $\pi : n \to m$, with $m \leq n$, and $y \in X_m$ with $x = \pi(y)$. One way to show what you want is through an argument with the skeleta; I'll give a more conceptual answer as to how we can calculate the geometric realization using just non-degenerate simplices. As an abuse of notation, for $\pi \in \Delta$ I will write $\pi(x)$ for the action on simplices and $pi(u)$ for the action on a point $u$ of a standard simplex under $|-| : \Delta \to \mathsf{Top}$
Recall that one way of defining the geometric realization functor is as the left Kan extension of $|-| : \Delta \to \mathsf{Top}$ along the Yoneda embedding $y : \Delta \to \mathsf{sSet}$. With this description, the geometric realization of $X \in \mathsf{sSet}$ is the colimit of $\int X \xrightarrow{\Pi} \Delta \xrightarrow{|-|} \mathsf{Top}$, where $\int X$ is the category of elements of $X$: objects are $(n,x)$ for $[n] \in \Delta$ and $x \in X_n$, and a morphism $(n,x) \to (m,y)$ is a map $\pi : [n] \to [m] \in \Delta$ such that $\pi(y)=x$ (see Category Theory in Context, definition 2.4.2). The functor $\Pi : \int X \to \Delta$ is the projection given by $\Pi(n,x) = [n]$ and $\Pi(\pi: (n,x) \to (m,y)) = \pi: [n] \to [m]$.
What we can use to show your proposition is the notion of a final functor (sometimes called cofinal). A functor $F : C \to D$ is final if, for any $d \in D$, the comma category $d \downarrow F$ is non-empty. That is, for any $d \in D$, there is some $c \in C$ and morphism $g : d \to Fc$, and moreover, any two morphisms are connected by a zigzag of commutative triangles. Final functors have the property that, for any diagram $G : D \to E$ which admits a colimit, the induced map $\operatorname{colim} GF \to \operatorname{colim} G$ is an isomorphism (see Categories for the Working Mathematician, Chapter IX.3).
Now, let $\int_{nd}X$ is the full subcategory of $\int X$ spanned by those $(c, x)$ for which $x$ is nondegenerate. Using the EZ-lemma and Eilenberg-Zilber axioms, you can show that the inclusion $\iota : \int_{nd} X \to \int X$ is final. Thus, we can calculate the geometric realization of a simplicial set $X$ as the colimit of
$$\int_{nd} X \to \int X \xrightarrow{\Pi} \Delta \xrightarrow{|-|} \mathsf{Top}$$
Call this composite $S_X$, so $S_X(n,x) = \Delta^n$; I will identify $S_X(n,x) \cong \{x\} \times \Delta^n$ to keep track of which simplex things are paired with. Explicitly, using some standard formulas for colimits, this can be calculated as the coequalizer of:
$$\coprod_{\sigma : (n,x) \to (m,y) \in \operatorname{Mor}\int_{nd} X} S_X(n,x) \rightrightarrows \coprod_{(k,z) \in \int_{nd} X}S_X(k,z)$$
where one of the parallel maps is induced by the cone $$\left\{S_X(n,x) \xrightarrow{S_X(\pi)} S_X(m,y) \hookrightarrow \coprod_{(k,z) \in \int_{nd} X} S_X(k,z)\right\}_{\pi: (n,x) \to (m,y)}$$ and the other is induced by the cone $$\left\{S_X(n,x) \xrightarrow{id} S_X(n,x) \hookrightarrow \coprod_{(k,z) \in \int_{nd} X} S_X(k,z)\right\}_{\pi: (n,x) \to (m,y)}$$
We may identify
$$\coprod_{(k,z) \int_{nd} X} S_X(k,z) \cong \coprod_{k \in \mathbb{N}} \overline{X_k} \times \Delta^k$$
with $\overline{X_n}$ denoting non-degenerate $n$-simplices, given the discrete topology. In this case, on the component corresponding to $\pi: (n,x) \to (m,y)$, our maps take $(x,v) \in S_X(n,x)$ to $(x, v)$ and $(y, \pi (v))$. Recalling our definition of the category of elements, here $x = \pi(y)$.
But we know how to calculate coequalizers in $\mathsf{Top}$: here we'll define $\sim$ to be the smallest equivalence relation for which $(\pi(y), v) = (x,v) \sim (y, \pi(v))$ for each $\pi : (n,x) \to (m,y)$, and then take $|X| = \operatorname{colim} S_X = \coprod_{(k,z) \in \int_{nd} X} \Delta^k/\sim$. Thus, we have formed the geometric realization using just non-degenerate simplices; the end product looks similar to what you had originally suggested but with some modifications, so I hope it's still of use.
One detail worth checking is that the only maps $\pi$ in $\int_{nd} X$ are injective (hence can be written as composites of face maps), as otherwise this contradicts that all simplices involved are nondegenerate. Try thinking about what this would correspond to geometrically, and the differences compared to the "naive" construction taking all simplices of $X$.
In general if $X$ is a simplicial set, any $n$-simplex $\sigma\in X_n$ corresponds to a map of simplicial sets $\Delta^n\to X$. This follows from the Yoneda lemma. After realization this leads to a map $\vert\sigma\vert\colon\Delta^n_\mathrm{top} = \vert\Delta^n\vert\to \vert X\vert$.
This map can alternatively described as follows. Recall that $\vert X\vert$ is the quotient of $\coprod_{n=0}^\infty X_n\times \Delta^n_\mathrm{top}$. Then $\vert\sigma\vert$ is the composite
$$\Delta^n_\mathrm{top}\cong \{\sigma\}\times \Delta^n_\mathrm{top}\subset X_n\times \Delta^n_\mathrm{top}\to \vert X\vert.$$
We identify $\Delta^1_\mathrm{top}$ with the unit interval via the homeomorphism
$$[0,1]\xrightarrow{\cong}\Delta^1_\mathrm{top},t\mapsto (1-t,t),$$
and we see in particular that any $1$-simplex $\sigma$ of $X$ gives rise to the path $\vert\sigma\vert$ with starting point $\vert d_1(\sigma)\vert$ and endpoint $\vert d_0(\sigma)\vert$. Also, $\vert\sigma\vert$ is constant if $\sigma$ is a degenerate $1$-simplex. Finally, note that if $\sigma$ is a $2$-simplex, then the composition of the paths $\vert d_2(\sigma)\vert$ and $\vert d_0(\sigma)\vert$ is path homotopic to $\vert d_1(\sigma)\vert$.
In the special case of $N(\mathcal{C})$ the $0$-simplices are the objects and the $1$-simplices are the morphisms. Note that if $f\colon x\to y$ is a morphism, then $d_1(f) = x$ and $d_0(f) = y$. Also, if $\sigma$ is the $2$-simplex $x\xrightarrow{f} y\xrightarrow{g} z$, then $d_0(\sigma) = g, d_1(\sigma) = g\circ f$ and $d_2(\sigma) = f$.
From the previous discussion we immediately see that we obtain a functor $\mathcal{C}\to \Pi_1(\vert N(\mathcal{C})\vert$.
Finally, if $M$ and $BM$ is the category with one object $\ast$ and a morphism for any element of $M$, then $\vert N(BM)\vert$ is a CW-complex with one vertex and an edge for every $m\in M\setminus\{1\}$. A $2$-simplex $\ast\xrightarrow{m}\ast\xrightarrow{n}$ is non-degenerate iff $m\neq 1\neq n$. We see that $\pi_1(\vert N(BM)\vert)$ is the free group with a generator $e_m$ for every $m\in M\setminus\{1\}$ modulo the relation $e_m\cdot e_n = e_{m\cdot n}$, which is exactly the group completion of $M$.
Best Answer
Just observe that there is a map $|X|\to |\Delta^n|$: each element of $X_k=\Delta(k,n)$ determines a map $|\Delta^k|\to|\Delta^n|$, and so these maps together over all values of $k$ give a map $\coprod_kX_k\times|\Delta^k|\to|\Delta^n|$. This map respects the equivalence relation $\sim$, essentially by definition (if two elements of $X_k\times|\Delta^k|$ and $X_{k+1}\times|\Delta^{k+1}|$ are related by a face or degeneracy, the corresponding maps $|\Delta^k|\to |\Delta^n|$ and $|\Delta^{k+1}|\to|\Delta^n|$ are related by composing with a map between $|\Delta^k|$ and $|\Delta^{k+1}|$ and this gives exactly the relation required by $\sim$), and thus descends to a map $f:|X|\to |\Delta^n|$.
If $i:|\Delta^n|\to |X|$ is the map given by the inclusion of $\{id_n\}\times|\Delta^n|$, then it is clear that $fi$ is the identity, and so in particular $i$ is injective. On the other hand, the work you have done proves exactly that $i$ is surjective. Thus $i$ is a bijection, and $f$ is its inverse, and so they are inverse homeomorphisms.