Consider the separable metric space $X=[0,1]\times\{0,1\}$ endowed with the $\ell_1$-metric $d:X\times X\to\mathbb R$ defined by $$d((x,i),(y,j))=|x-y|+|i-j|. $$ It seems that this metric space yields a counterexample to your question.

Consider the separable metric space $X=[0,1]\times\{0,1\}$ endowed with the $\ell_1$-metric $d:X\times X\to\mathbb R$ defined by $$d\big((x,i),(y,j)\big)=|x-y|+|i-j|. $$ It seems that this metric space yields a counterexample to your question.

Consider the separable metric space $X=[0,1]\times\{0,1\}$ endowed with the metric $d:X\times X\to\mathbb R$ defined by $$d((x,i),(y,j))=\begin{cases} |x-y|&\mbox{if $i=j$};\\ \sqrt{1+|x-y|}&\mbox{if $i\ne j$}. \end{cases} $$ It seems that this metric space yields a counterexample to your question.

Consider the separable metric space $X=[0,1]\times\{0,1\}$ endowed with the $\ell_1$-metric $d:X\times X\to\mathbb R$ defined by $$d((x,i),(y,j))=|x-y|+|i-j|. $$ It seems that this metric space yields a counterexample to your question.