Let be $X$ a set. So if $\delta$ and $\rho$ are two metrics on $X$ then we say that $\delta\preceq \rho$ if for any $\epsilon\in\Bbb R^+$ exists $\nu_\epsilon\in\Bbb R^+$ such that $$ \delta(x,y)\le\nu_\epsilon\cdot \rho(x,y) $$ when $x,y\in X$ are such that $\rho(x,y)<\epsilon$.
Now let be $\delta$, $\rho$ and $\sigma$ three metrics on $X$. So first of all we observe given any $\epsilon\in\Bbb R^+$ then any $\nu\in[1,+\infty)$ is such that $$ \delta(x,y)\le\nu\cdot\delta(x,y) $$ when $x,y\in X$ are such that $\delta(x,y)<\epsilon$ so that we conclude that the relation $\preceq$ is reflexive.
Moreover if $\delta\preceq\rho$ and $\rho\preceq\sigma$ then for any $\epsilon\in\Bbb R^+$ there exist $\nu_\epsilon$ and $\mu_\epsilon$ such that $$ \delta(x,y)\le\nu_\epsilon\cdot\rho(x,y)\,\,\,\text{and}\,\,\,\rho(x,y)\le\mu_\epsilon\cdot\sigma(x,y) $$ for any $x,y\in X$ whose distance is less than $\epsilon$ with respect $\rho$ and $\sigma$ and thus we conclude that $\nu_\epsilon\mu_\epsilon$ is a positive real number such that $$ \delta(x,y)\le\nu_\epsilon\mu_\epsilon\cdot\sigma(x,y) $$ for any $x,y\in X$ whose distance is less than $\epsilon$ with respect $\sigma$ so that we finally conclude that $\delta\preceq\sigma$, that is the relation $\preceq$ is transitive. So we can claim that the relation $\preceq$ is effectively a preorder, provided that the argumentations I gave are not incorrect.
Now unfortunately I was not able to prove that if $\delta\preceq\rho$ then the topology generated by $\delta$ is less finer than the topology generated by $\rho$, that is if $\delta\preceq\rho$ then why any open ball of $\delta$ is union of open balls of $\rho$? I tried to prove that if $\delta\preceq\rho$ then there exist a constant $\kappa$ such that $$ \delta(x,y)\le\kappa\cdot\rho(x,y) $$ for any $x,y\in X$ but unfortunately I failed. So could someone help me, please?
To follow the solution proposed by the professor Brian M. Scott in the comments below
So let be $\mathcal T_\delta$ and $\mathcal T_\rho$ the topology generated by $\delta$ and $\rho$. So if $A\in\mathcal T_\delta$ then for any $x_0\in A$ there exists $\epsilon\in\Bbb R^+$ such that $$ x_0\in B_\delta(x_0,\epsilon)\subseteq A $$ Now if $\delta\preceq\rho$ there exist $\nu_\epsilon$ such that $$ \delta(x,y)\le\nu_\epsilon\cdot\rho(x,y) $$ for any $x,y\in X$ whose distance with respect $\rho$ is less than $\epsilon$. So putting $$ \varepsilon:=\min\Big\{\epsilon,\frac\epsilon{\nu_\epsilon}\Big\} $$ we observe that $$ B_\rho(x_0,\varepsilon)\subseteq B_\rho(x_0,\epsilon) $$ so that we conclude that $$ \delta(x_0,x)\le\nu_\epsilon\cdot\rho(x_0,x)<\nu_\epsilon\cdot\varepsilon\le\nu_\epsilon\cdot\frac\epsilon{\nu_\epsilon}=\epsilon $$ for any $x\in B_\rho(x_0,\varepsilon)$ and thus we finally conclude that $$ x_0\in B_\rho(x_0,\varepsilon)\subseteq B_\delta(x_0,\epsilon)\subseteq A $$ which proves that $A$ is open in $\mathcal T_\rho$ and so $$ \mathcal T_\delta\subseteq\mathcal T_\rho $$
Now we suppose that $$ \delta\preceq\rho\quad\text{and}\quad\rho\preceq\sigma $$ So if $\delta\preceq\rho$ then for any $\epsilon\in\Bbb R^+$ there exists $\nu_\epsilon$ such that $$ \rho(x,y)\le\nu_\epsilon\cdot\sigma(x,y)<\nu_\epsilon\cdot\epsilon $$ for any $x,y\in X$ whose distance with respect $\sigma$ is less than $\epsilon$ but if $\delta\preceq\rho$ then there exists $\mu_{\nu_\epsilon}\in\Bbb R^+$ such that $$ \delta(x,y)\le\mu_{\nu_\epsilon}\cdot\rho(x,y) $$ for any $x,y\in X$ whose distance with respect $\rho$ is less than $\nu_\epsilon\cdot\epsilon$ so that finally we conclude that if $x,y\in X$ are distant with respect $\sigma$ by a smaller amount than $\epsilon$ then $$ \delta(x,y)\le\mu_{\nu_\epsilon}\cdot\rho(x,y)\le\big(\mu_{\nu_\epsilon}\cdot\nu_\epsilon\big)\cdot\sigma(x,y) $$ which proves that the relation $\preceq$ is transitive.