There are more than one way to prove the invariance of Euler characteristic under barycentric subdivision. A self-contained and clear solution can be found here: http://www.math.uni-frankfurt.de/~johannso/SkriptAll/SkriptTopGen/SkriptTopI/vorl2.pdf . However, if the solution is to do induction on the number of simplicies, I believe the following solution is what you asked for. It does not make use of any lemma regarding Euler characteristic, but only some observation.
After barycentric subdivision, the Euler characteristic is transformed into $\chi(\text{sd}(K))=\sum_{n=0}^{N} (-1)^n \alpha'_n$ where $\alpha'_n$ is the number of $n$-simplices in sd($K$). We observe that a $n$-simplex is formed by, given a $(n-1)$-simplex and a $n$-th point outside the hyperplane generated by those $(n-1)$ vertices, joining the $n$-th point to all those $(n-1)$ vertices. Since any vertex of sd$(K)$ is a barycentre of some simplex of $K$, any $n$-simplex in sd$(K)$ can be interpreted as having the $n$-th vertex as the barycentre of some $i$-simplex in $K$, and the other $(n-1)$ vertices as forming a $(n-1)$-simplex in that $i$-th simplex. Denote $K^p_q$ as, after barycentric subdivision, the number of $q$-simplices (of sd$(K)$) in a $p$-simplex (of $K$) that does not have any vertice being the barycentre of the $p$-simplex. Note that $q\leq p$ is required and $K^p_p=0$. Then for any $n$, $0< n \leq N$:
\begin{align*}
\alpha_n'=\alpha_n K^n_{n-1} +\alpha_{n+1} K^{n+1}_{n-1} +...+\alpha_N K^N_{n-1}
\end{align*}
It is clear that $\alpha_0'=\alpha_0+\alpha_1+...+\alpha_N$. Then:
\begin{align*}
\chi(\text{sd}(K))
&= \sum_{n=0}^{N} (-1)^n \alpha'_n\\
&= \alpha_0 + \alpha_1(1-K^1_0) +\alpha_2(1-K^2_0+K^2_1)+...+\alpha_N(1-K^N_0+K^N_1-...\pm K^N_{N-1})\\
&= C_0\alpha_0 + C_1\alpha_1 + C_2\alpha_2 +...+C_N\alpha_N
\end{align*}
where $\pm$ is $+$ if $N$ is even and $-$ if $N$ is odd. A quick check shows that $K^1_0=2$, so $C_1=1-K^1_0=-1=-C_0$. If we can show that for any $n\leq N$, $C_n=-C_{n-1}$, then $\chi'(\text{sd}(K))=\alpha_0-\alpha_1+\alpha_2-...\pm\alpha_N$ and we're done.
To do so, we carry out induction: with the hypothesis $K^{n+1}_q=(q+2)(K^n_q+K^n_{q-1})$ for $1\leq q\leq n-1\leq N-1$, we proceed to prove that $K^{n+1}_{q+1}=(q+3)(K^n_{q+1}+K^n_{q})$. Denote $k^r_s$ as the number of $s$-simplices in a $r$-simplex (before barycentric subdivision). By the same observation, the expression for $K_q^n$ is very similar to that of $\alpha_n'$ derived above:
\begin{align*}
K^n_q = k^n_qK^q_{q-1} + k^n_{q+1}K^{q+1}_{q-1}+...+k^n_{n-1}K^{n-1}_{q-1}
\end{align*}
Also, we note that $k^{n+1}_q=k^n_q+k^n_{q-1}$, and $k^n_n=1$. Plug in this relation and the induction hypothesis, a few steps of simple algebraic manipulation obtain the conclusion. By the same procedure, we can check that (using $K^{n+1}_0=2(K^n_0+1)$) $K^{n+1}_1=3(K^n_1+K^n_0)$, which can serve as the starting point for induction. Therefore $K^{n+1}_q=(q+2)(K^n_q+K^n_{q-1})$ for $1\leq q\leq n\leq N$.
Now, for $n$ even:
\begin{align*}
C_n &=1-K^n_0 + K^n_1 - K^n_2 +...-K^n_{n-2}+K^n_{n-1}\\
&= 1 - 2(K^{n-1}_0+1) + 3(K^{n-1}_1 + K^{n-1}_0) - 4(K_2^{n-1}+K_1^{n-1} \\+ ... &-n(K^{n-1}_{n-2}+K^{n-1}_{n-3}) + (n+1)(K^{n-1}_{n-1}+K^{n-1}_{n-2})\\
&= -1 + K^{n-1}_0 - K^{n-1}_1 + ... + K^{n-1}_{n-2}\\
&= -C_{n-1}
\end{align*}
The same result yields for $n$ odd, by explicit calculation like above. Therefore, the Euler characteristic is invariant under barycentric subdivision.
