Yes, such a distance $d'$ exists.
We can suppose $d(x,X)=d(y,X)=k$ for some $k>0$. We can define a new distance $d'$ by $d'(a,b)=d(a,b)+|d(x,a)-d(x,b)|$. This easily implies $d'(x,X)=2k$, however for $d'(y,X)$ to be $2k$ we would need a point $a\in X$ such that
$2k=d(y,a)+|d(x,a)-d(x,y)|$.
However $d(y,a),d(x,a)$ and $d(x,y)$ are all $\leq k$ and $d(x,y)>0$, so this implies $d(x,a)=0$, so $x=a$ and $d(x,y)=k$ so $d(x,y)=d(x,X)$. So this method works except if $d(x,y)=d(x,X)$.
But it is easy to give $X$ a metric such that $d(x,y)\neq d(x,X)$: for each $\varepsilon\in(0,d(x,y))$ define $r_\varepsilon:X^2\to[0,\infty)$ by $r_\varepsilon(x,y)=r_{\varepsilon}(y,x)=\varepsilon$ and $r_\varepsilon(a,b)=d(a,b)$ for any other pair of points of $X$, and define the distance
$d_\varepsilon(a,b)=\inf\{\sum_{i=1}^nr_\varepsilon(x_{i-1},x_i);(x_0,\dots,x_n)\text{ sequence of points of $X$ with }x_0=a,x_n=b\}$$d_\varepsilon(a,b)=\inf\{\sum_{i=1}^nr_\varepsilon(x_{i-1},x_i);x_0,\dots,x_n\text{ points of $X$ with }x_0=a,x_n=b\}$.
Note that if $d(a,b)<\varepsilon$, then $d_\varepsilon(a,b)=d(a,b)$, so $d_\varepsilon$ is a distance generating the same topology as $d$, and if $X$ has more points apart from $x$ and $y$, then for small enough $\varepsilon$ (specifically, take $\varepsilon<\min(d(x,z),d(y,z))$ for some $z\not\in\{x,y\}$) $d_\varepsilon(x,y)$ is less than $d_\varepsilon(x,X)$.