d' is finer than d on compact space $\implies$ $\forall \epsilon \ \ \exists \epsilon' \ \ \forall x \ B_{d'}(x,\epsilon') \subseteq B_d(x,\epsilon)$ This is my conjecture, but I guess I am missing the key idea for the proof (or my conjecture is wrong)
Let d and d' be two metrics on a compact space $X$ ($X$ is compact with respect to both metrics). If d' generates a finer topology than d on $X$, then for every $\epsilon >0$ there exists $\epsilon'>0$ such that for every $x \in X$ $$B_{d'}(x,\epsilon') \subseteq B_d(x,\epsilon)$$ holds. (So basically this $\epsilon'$ works for any $x \in X$)
My attempts are meaningless, I tried to use total boundedness, but it leads nowhere. So, can you please tell me if my conjecture is true or not, and if yes can you give me a proof or an idea for the proof?
Thanks for any help.
 A: Suppose that $(X,d')$ is compact (so $(X,d)$ is compact as well).
Given $\epsilon>0$, we have the cover $\left\{B_d(x,\frac{\varepsilon}{2}):x\in X\right\}$ of $X$ by $d'$-open sets. If $\epsilon'$ is a Lebesgue number for that cover, then $\epsilon'$ satisfies the condition you want.

Notice that the conjecture means that the identity $(X,d')\to (X,d)$ is uniformly continuous.
If we assume only $(X,d)$ to be compact (but not $(X,d')$), then this conjecture is not true, even if $(X,d')$ is complete.
For example, let $X=[-1,\infty)$ with the usual metric $d'$. Let $W$ be the Warsaw Circle (it's easier to understand it with the image). Let $f:[-1,\infty)\to W$ be a continuous, injective function such that $f(-1)=(0,1)$, $f(0)=(1,\sin 1)$ (so that $f$ covers the arc of $W$ in $[-1,0]$) and $f(x)=(e^{-x},\sin e^x)$ for $x>0$.
Consider the metric $d(x,y)=\Vert f(x)-f(y)\Vert_{\mathbb{R}^2}$ in $X$, so that $(X,d)$ is compact (because $f:(X,d)\to(W,\Vert\cdot\Vert_{\mathbb{R}^2})$ is a homeomorphism) and, since $f$ is continuous, then it follows easily that the usual metric (topology) $d'$ is finer than $d$, but the identity $(X,d')\to(X,d)$ is not uniformly continuous.
