$d^\prime \le c\,d$ where $d,d^\prime$ are bounded metrics Let $X$ be a nonempty set. Fix two metrics $d: X\times X \to [0,1]$ and $d^\prime: X\times X \to [0,1]$ such that the topology $\tau$ generated by $d$ is finer than the topology $\tau^\prime$ generated by $d^\prime$, i.e.,
$
\tau^\prime \subseteq \tau.
$

Question. Is it true that there exists a positive constant $c$ such that 
  $$
\forall x,y \in X,\,\,\,\,\,\,\,d^\prime(x,y) \le c\,d(x,y)\,\,?
$$

 A: No. Here is an example where the metrics even induce the same topology:
Let $X=2^\omega$ (i.e., all functions $\mathbb N\rightarrow\{0,1\}$) where the space is endowed with the product topology where $\{0,1\}$ has the discrete topology (this is the topology generated by the basic open sets $N_s=\{x\in 2^\omega\mid s\subset x\}$ where $s\in 2^{<\omega}$, i.e., the basic open sets consist of all sequences extending one given finite string). This space is homeomorphic to the Cantor set and is called Cantor space.
The topology is generated by the metric
$$d:X\times X\rightarrow [0,1], (x,y)\mapsto\begin{cases}0\, ,&\text{ if }x=y,\\
\frac 1{n+1}\,, &\text{ otw., where $n$ is maximal s.t. }x\upharpoonright n=y\upharpoonright n\,.\end{cases}$$
This is obviously symmetric, has range $\subseteq [0,1]$, and satisfies the triangle inequality (just note that if $x$ and $y$ agree up to $n$ and $y$ and $z$ agree up to $k$, then $x$ and $z$ agree up to $\min\{k,n\}$).
But, the same topology is generated by the metric
$$d:X\times X\rightarrow [0,1], (x,y)\mapsto\begin{cases}0\, ,&\text{ if }x=y,\\
\frac 1{2^n}\,, &\text{ otw., where $n$ is maximal s.t. }x\upharpoonright n=y\upharpoonright n\,.\end{cases}$$
It is also easy to see that this is a metric.
Quite obviously, these metrics are not equivalent (since the exponential term gets bigger much faster than the linear term).
A: No, it is not true, and here is a counterexample: Let $X$ be the subset $\{1/n\mid n\in \Bbb N\}\subseteq\Bbb R$ and let $d$ be the metric inherited from the usual metric on the real line. Then $\tau$ is the discrete topology, and therefore the finest one there is.
Let $f:X\to \Bbb Q$ be a bijection, and let $d'$ be defined by $d'(x,y)=\min(1,|f(x)-f(y)|)$. This makes $(X,\tau')$ homeomorphic to $\Bbb Q$, and therefore not discrete.
So we have $\tau'\subsetneq\tau$. Assume there is such a $c$, and look at $Y=f^{-1}(\Bbb Z)$. The $d'$-distance between any two points in $Y$ is $1$, which means that the $d$-distance between any such points is greater than $\frac1c$. This cannot be, since only finitely many points of $X$ are above $\frac1c$ as points in $\Bbb R$, and $Y$ contains infinitely many points.
A: 
Suppose that your claim were true and that, in addition to your hypothesis:

*

*$(X,d')$ is a non-compact bounded metric space such that $\overline{(X,d')}$ is compact

*$(X,d)$ is a bounded metric space which is not relatively compact in $\overline{(X,d)}$.

Where $\overline{(Y,\delta)}$ is the completion of the metric space $(Y,\delta)$.
By your lemma, the map $id_X: (X,d')\to (X,d)$ is UC. By the universal property of the completion, there is exactly one uniformly continuous map $j:\overline{(X,d')}\to\overline{(X,d)}$ such that $j\rvert_X=id_X$. But this cannot be, because $\operatorname{im}j$ would be a compact subset of $\overline{(X,d)}$ containing $X$.

Now, the question is: can such a counterexample be produced under the hypothesis $\tau'\subseteq \tau$?
The first counterexample that comes to mind is with $\tau=\tau'$. $\Bbb R$ is homeomorphic to $S^1\setminus \{-1\}$ through the map $$\Bbb R\ni x\mapsto\psi(x)=e^{2i\arctan x}\in S^1\setminus\{-1\}$$
And thus it inherits a bounded distance $d'(x,y)=\lvert \psi(x)-\psi(y)\rvert$ such that $\overline{(\Bbb R,d')}\cong S^1$.
However, $\Bbb R$ also has the bounded distance $d(x,y)=\min\{1,\lvert x-y\rvert\}$, with repsect to which it is already its own completion (and it is not compact).
Added: Another one would be with the $1-0$ distance on $\Bbb R$, in which case $\tau'\subsetneq \tau$.
