We say that $\tau$ is finer than $\tau'$ if every open set of of $\tau'$ is open in $\tau$.
The question, if so, simply asks to show that every open set of $X$ is open in $X'$. Of course it is enough to show this for basic open sets, in our case $B(x,\epsilon)=\{y\mid d(x,y)<\epsilon\}$ sort of sets.
Fix $x\in X'$ and use the assumption that $f(y)=d(x,y)$ is a continuous function on $X'$ to deduce that $B(x,\epsilon)$ is open in $X'$.
To show that the $d$-metric topology is coarser than $\mathcal{T}$ we must show that every $d$-open set is $\mathcal{T}$-open. Of course, it really suffices to show that every $d$-ball is $\mathcal{T}$-open (since the $d$-balls form a basis for the $d$-metric topology). To show that the $d$-ball $B ( x , \epsilon )$ is $\mathcal{T}$-open it suffices to find for each $y \in B ( x , \epsilon )$ a $\mathcal{T}$-open neighbourhood $V$ of $y$ such that $V \subseteq B ( x , \epsilon )$.
First, the continuity of $d$ implies that for each $\epsilon > 0$ the set $$U_\epsilon := \{ ( u , v ) \in X \times X : d ( u , v ) < \epsilon \} = d^{-1} [ ( -\infty , \epsilon ) ]$$ is open in $X \times X$ with respect to the topology $\mathcal{T}$ (or $\mathcal{T} \times \mathcal{T}$, if you will).
Now, given $x \in X$ and $\epsilon > 0$, we want to show for all $y \in B ( x , \epsilon )$ that there is a $\mathcal{T}$-open set $V$ with $y \in V \subseteq B ( x , \epsilon )$. As $y \in B ( x , \epsilon )$, it follows that $( x , y ) \in U_\epsilon$, and since $U_\epsilon$ is a $( \mathcal{T} \times \mathcal{T} )$-open subset of $X \times X$, there are $\mathcal{T}$-open $U , V \subseteq X$ such that $$( x , y ) \in U \times V \subseteq U_\epsilon.$$ In particular $V$ is a $\mathcal{T}$-open neighbourhood of $y$.
Note that given any $z \in V$ we have that $( x , z ) \in U \times V \subseteq U_\epsilon$, and so by definition of $U_\epsilon$ it follows that $d ( x , z ) < \epsilon$, meaning that $z \in B ( x , \epsilon )$. We may then conclude that $V \subseteq B ( x , \epsilon )$.
Best Answer
$(1)$ Note that $X'$ is only a topological space, not necessarily a metric space, and $d$ is only a continuous map, not necessarily a metric inducing the topology on $X'$ (even though in this case it is a metric because $X'$ has the same underlying set as $X$, the topology on $X'$ may be different than the topology defined by the metric $d$).
$(2)$ We are given that $X$ is a metric space in the problem but $X'$ isn't necessarily a metric space (although it may be) - it is only apriori a topological space. The metric on $X$ is still $d$, and in this case the metric is used to define the topology on $X$.
$X'$ may be a metric space but it's possible that, even if it is, the metric isn't given by $d$. For instance if $X$ is $\{\frac{1}{n}\mid n\in\mathbb{N}^{+} \}\cup\{0\}$ with the induced metric $d$ from $\mathbb{R}$, then we can give $X'$ the discrete metric $d'$, and $d\colon X'\times X'\to\mathbb{R}$ would still be a continuous map even though it is a different metric to $d$. (This idea of using the discrete metric for $X'$ works for any non-discrete metric space $X$). To see that these spaces really are different, note that $\{0\}\subset X$ is not an open set but $\{x\}\subset X'$ is open for all $x\in X'$ - that is, $X$ is not discrete but $X'$ is.