Having infinite metric dimension is not bi-Lipschitz-invariant. On the real line, consider the two metrics $$ d(x,y)=\min(|x-y|,1)\quad\text{and}\\ \delta(x,y)=\arctan(|x-y|). $$ The two metrics both give the usual topology, they are bi-Lipschitz to each other, but $\dim(\mathbb R,d)=\infty$ (as observed by Benoît Kloeckner) and $\dim(\mathbb R,\delta)=1$ (the $\delta$-distances from any two points uniquely determine the point).

Usually one might argue that if two metric spaces are bi-Lipschitz, they are essentially the same. However, this does not guarantee that the metric dimensions coincide. I don't know what would characterize infinite metric dimension, but it has to be "finer than Lipschitz properties" of the metric. Perhaps more structure emerges if we restrict ourselves to length metrics?

On a tangential end note, consider the metric $d_\epsilon(x,y)=\min(|x-y|,\epsilon)$ on $S^1$ for any $\epsilon>0$. Here the absolute value can be the metric inherited from $\mathbb R^2$ or $\mathbb R/2\pi\mathbb Z$, it doesn't really matter. All these are bi-Lipschitz to each other, but the dimension for $d_\epsilon$ is roughly $2\pi/\epsilon$. You can adjust the metric dimension of $S^1$ to be any positive integer with bi-Lipschitz changes.

Having infinite metric dimension is not bi-Lipschitz-invariant. On the real line, consider the two metrics $$ d(x,y)=\min(|x-y|,1)\quad\text{and}\\ \delta(x,y)=\arctan(|x-y|). $$ The two metrics both give the usual topology, they are bi-Lipschitz to each other, but $\dim(\mathbb R,d)=\infty$ (as observed by Benoît Kloeckner) and $\dim(\mathbb R,\delta)=1$ (the $\delta$-distances from any two points uniquely determine the point).

Usually one might argue that if two metric spaces are bi-Lipschitz, they are essentially the same. However, this does not guarantee that the metric dimensions coincide. I don't know what would characterize infinite metric dimension, but it has to be "finer than Lipschitz properties" of the metric. As Benoît Kloeckner's new answer indicates, it does not help if we restrict ourselves to length metrics.

On a tangential end note, consider the metric $d_\epsilon(x,y)=\min(|x-y|,\epsilon)$ on $S^1$ for any $\epsilon>0$. Here the absolute value can be the metric inherited from $\mathbb R^2$ or $\mathbb R/2\pi\mathbb Z$, it doesn't really matter. All these are bi-Lipschitz to each other, but the dimension for $d_\epsilon$ is roughly $2\pi/\epsilon$. You can adjust the metric dimension of $S^1$ to be any positive integer with bi-Lipschitz changes.

Having infinite metric dimension is not bi-Lipschitz-invariant. On the real line, consider the two metrics $$ d(x,y)=\min(|x-y|,1)\quad\text{and}\\ \delta(x,y)=\arctan(|x-y|). $$ The two metrics both give the usual topology, they are bi-Lipschitz to each other, but $\dim(\mathbb R,d)=\infty$ (as observed by Benoît Kloeckner) and $\dim(\mathbb R,\delta)=1$ (the $\delta$-distances from any two points uniquely determine the point).

Usually one might argue that if two metric spaces are bi-Lipschitz, they are essentially the same. However, this does not guarantee that the metric dimensions coincide. I don't know what would characterize infinite metric dimension, but it has to be "finer than Lipschitz properties" of the metric.

On a tangential end note, consider the metric $d_\epsilon(x,y)=\min(|x-y|,\epsilon)$ on $S^1$ for any $\epsilon>0$. Here the absolute value can be the metric inherited from $\mathbb R^2$ or $\mathbb R/2\pi\mathbb Z$, it doesn't really matter. All these are bi-Lipschitz to each other, but the dimension for $d_\epsilon$ is roughly $2\pi/\epsilon$. You can adjust the metric dimension of $S^1$ to be any positive integer with bi-Lipschitz changes.

Having infinite metric dimension is not bi-Lipschitz-invariant. On the real line, consider the two metrics $$ d(x,y)=\min(|x-y|,1)\quad\text{and}\\ \delta(x,y)=\arctan(|x-y|). $$ The two metrics both give the usual topology, they are bi-Lipschitz to each other, but $\dim(\mathbb R,d)=\infty$ (as observed by Benoît Kloeckner) and $\dim(\mathbb R,\delta)=1$ (the $\delta$-distances from any two points uniquely determine the point).

Usually one might argue that if two metric spaces are bi-Lipschitz, they are essentially the same. However, this does not guarantee that the metric dimensions coincide. I don't know what would characterize infinite metric dimension, but it has to be "finer than Lipschitz properties" of the metric. Perhaps more structure emerges if we restrict ourselves to length metrics?

On a tangential end note, consider the metric $d_\epsilon(x,y)=\min(|x-y|,\epsilon)$ on $S^1$ for any $\epsilon>0$. Here the absolute value can be the metric inherited from $\mathbb R^2$ or $\mathbb R/2\pi\mathbb Z$, it doesn't really matter. All these are bi-Lipschitz to each other, but the dimension for $d_\epsilon$ is roughly $2\pi/\epsilon$. You can adjust the metric dimension of $S^1$ to be any positive integer with bi-Lipschitz changes.