Suppose that $m_1$ and $m_2$ are two metrics on a set $X$. I want to prove that if there exist positive constants $d,c,C$ such that $m_1(x,y)\leq d$ implies $cm_1(x,y)\leq m_2(x,y)\leq Cm_1(x,y)$, then $m_1$ and $m_2$ are equivalent.
But how should the constant $d$ be used to prove this claim? With asuming the inequality above alone, the proof is fairly straightforward.
The statement is false. Let $X=\mathbb{R}$. Let $m_1$ be the discrete metric, where $m_1(x,y)$ is zero if $x=y$, and one otherwise. Let $m_2$ be the standard Euclidean metric $m_2(x,y)=|x-y|$.
Choose $d=0.5, c=C=1$. Now, $m_1(x,y)\le d$ implies that $x=y$, and hence each of $cm_1(x,y), m_2(x,y), Cm_1(x,y)$ is zero. Your condition holds. However, $m_1$ and $m_2$ are not equivalent. Let $x=0, y\to \infty$; $m_2(x,y)$ grows without bound, and must be larger than $Cm_1(x,y)=C$ for sufficiently large $y$.