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From

If you take in (1) it follows that $x=z=w$ and then you get $$d(tx+(1-t)y,x)\leq (1-t){\cdot}d(x,y),$$ the same way you get $$d(tx+(1-t)y,y)\leq t {\cdot}d(x,y).$$ By the triangle inequality, you have "=" in the both of these inequalities. I.e., $$d(tx+(1-t)y,x)= (1-t){\cdot}d(x,y).$$ Applying it twice, you get equality in (2) with $C=d(x,y)$. Thus I.e., your metric is induced by a norm.

So any finite dimensional subspace is bi-Lipschitz to Euclidean space. Hence you get that virtually all dimensions of your space coincide.

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It seems that your conditions imply

From (1) it follows that you have equality in (2) with $C=d(x,y)$. Thus your metric is induced by a norm(?) [(1) implies that in (2) you have $C=d(x,y)$.].

So any finite dimensional subspace is bi-Lipschitz to Euclidean space. Hence you get that virtually all dimensions of your space coincide.

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It seems that your conditions imply that your metric is induced by norm (?) [(1) implies that in (2) you have $C=d(x,y)$.]

So any finite dimensional subspace is bi-Lipschitz to Euclidean space. Hence you get that virtually all dimensions of your space coincide.