$\min (d,1)$ and $d\mathrm{/}(1+d)$ are metrics uniformly equivalent to $d$, so every metric space carries a bounded metric with the same topology

Properties of ${\phi }_1$ on $t\ge 0$: it is one of $t$ and $1$, so $0\le {\phi }_1(t)\le t$ and ${\phi }_1(t)\le 1$; ${\phi }_1(t)=0$ exactly when $t=0$, since $1\mathrm{\ne }0$; it is nondecreasing, because for $0\le s\le t$ the value ${\phi }_1(s)$ is $s\le t$ or $1$, and in both cases it is a lower bound of $\{t,1\}$, hence at most ${\phi }_1(t)$; and it is subadditive, since for $s,t\ge 0$ either one of $s,t$ is $\ge 1$, and then ${\phi }_1(s)+{\phi }_1(t)\ge 1\ge {\phi }_1(s+t)$, or both are $<1$, and then ${\phi }_1(s)+{\phi }_1(t)=s+t\ge {\phi }_1(s+t)$.

Properties of ${\phi }_2$ on $t\ge 0$: here $1+t\ge 1>0$, so $(1+t{)}^{-1}>0$ and $(1+t{)}^{-1}\le 1$, whence $0\le {\phi }_2(t)\le t$; also $t<1+t$ gives ${\phi }_2(t)<(1+t)(1+t{)}^{-1}=1$; ${\phi }_2(t)=0$ exactly when $t=0$; ${\phi }_2$ is strictly increasing, because ${\phi }_2(t)=1-(1+t{)}^{-1}$ and $0\le s<t$ gives $0<1+s<1+t$, hence $(1+t{)}^{-1}<(1+s{)}^{-1}$; and it is subadditive, since for $s,t\ge 0$ one has $0<1+s\le 1+s+t$ and $0<1+t\le 1+s+t$, so ${\phi }_2(s+t)=s(1+s+t{)}^{-1}+t(1+s+t{)}^{-1}\le s(1+s{)}^{-1}+t(1+t{)}^{-1}={\phi }_2(s)+{\phi }_2(t)$.

Both $d^{\mathrm{\prime }}$ and $d^{\mathrm{\prime }\mathrm{\prime }}$ are symmetric, being ${\phi }_i$ applied to the symmetric function $d$, and both vanish exactly on the diagonal, since ${\phi }_i(t)=0$ exactly when $t=0$ and $d(x,y)=0$ exactly when $x=y$.

$d^{\mathrm{\prime }}$ is a metric: (M1) and (M2) are step 1.3, and (M3) follows because $d(x,z)\le d(x,y)+d(y,z)$ with ${\phi }_1$ nondecreasing and subadditive on nonnegatives gives $d^{\mathrm{\prime }}(x,z)={\phi }_1(d(x,z))\le {\phi }_1(d(x,y)+d(y,z))\le {\phi }_1(d(x,y))+{\phi }_1(d(y,z))=d^{\mathrm{\prime }}(x,y)+d^{\mathrm{\prime }}(y,z)$.

$d^{\mathrm{\prime }\mathrm{\prime }}$ is a metric: identically, using that ${\phi }_2$ is increasing and subadditive on nonnegatives, $d^{\mathrm{\prime }\mathrm{\prime }}(x,z)\le {\phi }_2(d(x,y)+d(y,z))\le d^{\mathrm{\prime }\mathrm{\prime }}(x,y)+d^{\mathrm{\prime }\mathrm{\prime }}(y,z)$.

Boundedness: $d^{\mathrm{\prime }}(x,y)\le 1<2$ and $d^{\mathrm{\prime }\mathrm{\prime }}(x,y)<1<2$ for all $x,y$, so if $X\mathrm{\ne }\mathrm{\varnothing }$ then fixing any $x_0\in X$ gives $X\subseteq B_{d^{\mathrm{\prime }}}(x_0,2)$ and $X\subseteq B_{d^{\mathrm{\prime }\mathrm{\prime }}}(x_0,2)$, while $X=\mathrm{\varnothing }$ is bounded outright; and $1$ is an upper bound of all the distances, so $\mathrm{diam}(X)\le 1$ in both metrics when $X\mathrm{\ne }\mathrm{\varnothing }$. This is claim 2.

$d^{\mathrm{\prime }}$ is uniformly equivalent to $d$: given $\epsilon >0$, take $\delta :=\epsilon$, so that $d(x,y)<\delta$ gives $d^{\mathrm{\prime }}(x,y)\le d(x,y)<\epsilon$; and take ${\delta }^{\mathrm{\prime }}:=\min \{\epsilon ,1\}>0$, so that $d^{\mathrm{\prime }}(x,y)<{\delta }^{\mathrm{\prime }}\le 1$ forces ${\phi }_1(d(x,y))\mathrm{\ne }1$, hence $d^{\mathrm{\prime }}(x,y)=d(x,y)$ by [L2], hence $d(x,y)<{\delta }^{\mathrm{\prime }}\le \epsilon$.

$d^{\mathrm{\prime }\mathrm{\prime }}$ is uniformly equivalent to $d$: given $\epsilon >0$, take $\delta :=\epsilon$, so that $d(x,y)<\delta$ gives $d^{\mathrm{\prime }\mathrm{\prime }}(x,y)\le d(x,y)<\epsilon$; and take ${\delta }^{\mathrm{\prime }}:={\phi }_2(\epsilon )>0$, so that $d^{\mathrm{\prime }\mathrm{\prime }}(x,y)<{\delta }^{\mathrm{\prime }}$ forces $d(x,y)<\epsilon$, since $d(x,y)\ge \epsilon$ would give ${\phi }_2(d(x,y))\ge {\phi }_2(\epsilon )={\delta }^{\mathrm{\prime }}$ by monotonicity.

Uniform equivalence implies topological equivalence, so $d^{\mathrm{\prime }}$ and $d^{\mathrm{\prime }\mathrm{\prime }}$ have exactly the metric topology of $d$; this completes claim 3.

Claims 1, 2 and 3 hold by steps 2.1 and 2.2, step 2.3, and steps 3.1, 3.2 and 4.1; hence every metric space carries a bounded metric inducing the same topology.