First, let us fix some definitions.
Definition: an $n$-simplex is a $n$-dimensional polytope which is the convex hull of it's $n+1$ vertices. Importantly, no vertex is contained in the convex hull of any other vertices.
Definition: a simplicial complex $K$ is a set of simplicies such that any face of a simplex of $K$ is also in $K$ and the intersection of any two simplicies $\sigma_1, \sigma_2$ is a face of $\sigma_1$ and $\sigma_2$.
Part (a): In order to show that these vertices actually form the vertices of simplex, we must check that no vertex is contained in the convex hull of any of the others. Suppose that $b_k=(x_{0,k},x_{1,k},\cdots,x_{n,k})$ (in $v_i$ coordinates) is contained in the convex hull of the other $b_j$. This would mean that there's a nontrivial linear dependence relation on the the set $\{b_j\}$.
But if we have some linear relation $\sum_{i=0}^n a_ib_i=0$, we can replace the $b_i$ by their defining linear combinations $\frac{1}{n+1}(v_0+\cdots+v_n)$ and obtain a linear relation on the $v_i$. But this is clearly ridiculous as the vectors $v_i$ are linearly independent since we assumed they were the vertices of a simplex. Some quick calculation shows that if the coefficients of $v_i$ are all zero, then so too are the coefficients $a_i$.
Part (b): For this, we need to check that the union of these simplicies is the entire space and that any two simplicies intersect in simplices.
To do the first, I'll give a method for determining which simplex a given point lies in. Represent an arbitrary point in our simplex as the linear combination of $v_i$, ie $x=a_0v_0+\cdots+a_nv_n$. I claim that it lies in the simplex determined by $b_{\sigma_0},\cdots,b_{\sigma_n}$ where $\sigma_i$ is the simplex defined by taking the greatest $i+1$ elements of the set $\{v_j\}$ given an order by $v_i\geq v_j$ if $a_i\geq a_j$ and ties broken by the lexographic ordering. Note that this always returns an answer- so every point in our original simplex is in at least one simplex in the barycentric subdivision.
Next, we need to show that if any collection of simplicies intersect, they intersect in a simplex. Since a simplex is determined entirely by its vertices, this means that the intersection set is determined by intersection of the vertices of the simplicies in common. But this means that the intersection of simplicies is exactly a simplex with vertices which are the intersection of the vertex sets of the simplicies we're intersecting. (Since these form a subset of the vertices of a simplex, they again determine a simplex.)
Part (d): Apply (b) to see that the barycentric subdivision forms a simplicial complex. The geometric realizations are the same because the behavior on the barycentric subdivision is still totally determined by the behavior of the points that were inherited from the original simplex. This is a moral-of-the-story answer, because I'm not sure what definition of geometric realization you're working with (the one I know is defined as a functor from simplicial sets to compactly generated hausdorff topological spaces- if you post your definition and you're still interested, I can add more to this.)
Finally, I've noticed from your recent questions that you seem to be in the midst of self-studying a lot of the foundations for algebraic topology. I highly recommend actually taking a course in the subject and discussing issues with your professor and coursemates- live interaction with peers and teachers is a much more reliable and useful tool than consulting stackexchange every time you have a potential issue.
Let $Y$ be any simplicial set and assume we have two maps $f,g : \displaystyle\coprod_{x: \Delta^n \to X}\Delta^n \to Y$ that make the above diagram commute, that is $f\circ (id, j) = g\circ (id, i)$ .
I will use the following convention : an element of the coproduct $\coprod_{i\in I}X_i$ is denoted $(x,i)$, where $x\in X_i$
Now assume $h: X\to Y$ makes the whole thing commute. Let $x\in X_n$ be an $n$-simplex. Then $x$ corresponds to some $\tilde{x} : \Delta^n \to X$ (by the Yoneda lemma), and so $h_n(x) = f((id_{[n]},\tilde{x}))(=g((id_{[n]},\tilde{x}))$ by commutation of the diagram). Therefore if $h$ exists it is unique.
Now define $h$ degree-wise as above, with $f$ : it is well-defined on each degree by the Yoneda lemma. We must show that it is a simplicial map and that it makes the diagram commute, after this by the uniqueness above we will be done. That it makes the diagram commute is quite obvious, as $(id,i),(id,j)$ are surjective, so since $f\circ (id,j) = g\circ (id,i)$ and $h$ is defined through $f$, so this is clear.
Let's now prove that it is a simplicial map. Let $\varphi : [m]\to [n]$ be nondecreasing and $x\in X_n$. Let me write the action of $\varphi$ on the right, as $X$ is a contravariant functor of $[k]$.
Then $h(x\cdot \varphi) = f((id_{[n]},\widetilde{x\cdot\varphi})$, but $\widetilde{x\cdot \varphi}$ is nothing but $\tilde{x}\circ \overline{\varphi}$ where $\overline{\varphi}$ is the induced map $\Delta^m\to \Delta^n$; for this you have to explicit the Yoneda isomorphism.
Also, $(id_{[n]},\tilde{x}\circ \overline{\varphi}) = (id,j)(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x})$ where $y = x\cdot \varphi$, therefore $h(x\cdot \varphi) = g((id,i)(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x})$.
Now $i(id_{[m]}, \overline{\varphi}: \tilde{y}\to\tilde{x}) = (\overline\varphi(id_{[m]}), \tilde{x})$ by definition.
Now if you recall the definition of the induced map $\Delta^m\to \Delta^n$ and of the simplicial structure on $\Delta^n$, you see that $\overline\varphi (id_{[m]}) = id_{[n]}\cdot \varphi$. Therefore $h(x\cdot \varphi) =g((id_{[n]}, \tilde{x})\cdot \varphi) = g((id_{[n]},\tilde{x}))\cdot \varphi = h(x)\cdot \varphi$ : $h$ is simplicial; and we are done.
Best Answer
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 $X$.