I suppose this should be tagged as homework, since this is problem (1) on http://kurser.math.su.se/file.php/853/Top11.pdf.
The following might be of some help. I am fairly certain that it is correct. Let $K_n$ denote the simplicial complex consisting of a simplex of dimension $n$, together with all its faces. Let $K_n'$ denote its barycentric subdivision. Let $s(n)$ be the number of simplexes in $K_n$. Let $s(m,n)$ be the number of $m$-simplexes in $K_n$. Let $s'(m,n)$ be the number of $m$-simplexes in $K_n'$. Let for $n > 0$ $s''(m,n)$ be the number of $m$-simplexes in $K_n'$ not spanned by the barycenter of the $n$-simplex. Then
$$
s(n) = 2^{n+1}-1,
$$
and
$$
s(m,n) = \binom{n+1}{m+1},
$$
and
$$
s'(m,n) =
\begin{cases}
s(n) &\text{if } m=0, \\
s''(m,n) + s''(m-1,n) &\text{otherwise},
\end{cases}
$$
and
$$
s''(m,n) =
\begin{cases}
s(n)-1 &\text{if } m=0, \\
\sum\limits_{i=m}^{n-1}\left( s(i,n) \left[s'(m,i) - s''(m,i) \right] \right)
&\text{if } 0 < m < n, \\
0 &\text{if } m = n.
\end{cases}
$$
It is possible that the formulas could be considerably simplified. I can provide derivations of the above after homework deadline at May 13.
UPDATE: The expressions for $s(n)$ and $s(m,n)$ are stated at http://en.wikipedia.org/wiki/Simplex.
For $s'(m,n)$ reason as follows. The case $m = 0$ is clear. For $m > 0$, in addition to the $s''(m,n)$ $m$-simplexes not spanned by the barycenter of the $n$-simplex, there are $s''(m-1,n)$ $m$-simplexes spanned by the barycenter and the vertices of an $(m-1)$-simplex not spanned by the barycenter. There are no other $m$-simplexes.
For $s''(m,n)$ reason as follows. The cases $m = 0$ and $m = n$ are clear. For $0 < m < n$ and $m \leq i \leq n-1$, there are $s(i,n)$ $i$-complexes in $K_n$, each contributing to $K_n'$ with $s'(m,i)$ $m$-simplexes not spanned by the barycenter of the $n$-simplex. There are no other contributions of such simplexes. In the summation, discount for $m$-simplexes already contributed by simplexes of lower dimension: of the $s'(m,i)$ $m$-simplexes contributed by an an $i$-simplex, $s''(m,i)$ $m$-simplexes have already been counted (since these are not spanned by the barycenter of the $i$-simplex).
HOWEVER: Visualizing the complexes $K_n'$ for $n=0,1,2,3$, I get
$$
s'(0,0)=1 \\
s'(0,1)=3, \quad s''(0,1)=2, \\
s'(1,1)=2, \quad s''(1,1)=0, \\
s'(0,2)=7, \quad s''(0,2)=6, \\
s'(1,2)=12, \quad s''(1,2)=6, \\
s'(2,2)=6, \quad s''(2,2)=0, \\
s'(0,3)=15, \quad s''(0,3)=14, \\
s'(1,3)=50, \quad s''(1,3)=36, \\
s'(2,3)=?, \quad s''(2,3)=24, \\
s'(3,3)=24, \quad s''(3,3)=0.
$$
Using the formulas above, I get differing results $s'(1,3)=68$ and $s''(1,3)=54$ (and $s'(2,3)=78$). It seems the expression for $s''$ above is counting some contributions more than once (or I am counting simplexes wrong when visualizing $K_3'$).
UPDATE 2: Formula updated, gives correct results at least for $n \leq 3$ now. I think it is correct for all $n$ now.
Best Answer
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.