This is not true.
If $c(x,z)\le c(x,y) + 100\cdot c(y,z)$ , then $c(x,z)\le c(x,y) + c(y,z)$
However it is true that, for $x,z \in V$ with $x \neq z$, you have $c(x,z) \le 100 d(x,z)$.
To see this let $P$ be a shortest path from $x$ to $z$. The proof is by induction on the number of edges $|P|$ of $P$.
If $|P|=1$ then $c(x,z) = d(x,z) \le 100 d(x,z)$.
If $|P| \ge 2$ then, let $y$ be the vertex immediately preceding $z$ in $P$.
By inductive hypothesis $c(x, y) \le 100 d(x,y)$ and you have
$$
c(x,z) \le c(x,y) + 100 c(y,z) \le 100 d(x,y) + 100 c(y,z) =100 d(x,z). \quad \square
$$
Consider a MST $M$ of $G$ and let $T$ be a Euler tour of $M$, starting from an arbitrary vertex.
Let $v_0, v_1, \dots, v_{n-1}$ be the vertices in $V$ in the order they appear for the first time in $T$ and let $v_n = v_0$. Consider the tour $T'$ that visits $v_0, v_1, \dots, v_n$ in this order.
For $i=0, \dots, n-1$ let $T_i$ be the portion of the tour between the first occurrence of $v_i$ in $T$ and the next occurrence of $v_{i+1}$ in $T$.
Given any graph $H$, let $c(H)$ be the sum of costs of the edges in $H$.
Let $T^*$ an optimal tour of $G$.
You have:
$$
\begin{align*}
c(T') &= \sum_{i=0}^{n-1} c(v_i, v_{i+1}) \le 100 \sum_{i=0}^{n-1} d(v_i, v_{i+1}) \le 100 \sum_{i=0}^{n-1} c(T_i) \\
&\le 100 c(T) \le 200 c(M) \le 200 c(T^*).
\end{align*}
$$
