Suppose that $d_0$ and $d_1$ are metrics on a set $X$. Each generates a topology: $d_0$ generates $\tau_0=\left\{V\subseteq X:\forall x\in V\exists \epsilon_x>0\big(B_{d_0}(x,\epsilon_x)\subseteq V\big)\right\}\;,$ and $d_1$ generates $\tau_1=\left\{V\subseteq X:\forall x\in V\exists \epsilon_x>0\big(B_{d_1}(x,\epsilon_x)\subseteq V\big)\right\}\;,$ where $B_{d_0}(x,\epsilon)=\{y\in X:d_0(x,y)<\epsilon\}$ and $B_{d_1}(x,\epsilon)=\{y\in X:d_1(x,y)<\epsilon\}$. We say that $d_0$ and $d_1$ are topologically equivalent iff $\tau_0=\tau_1$: the two metrics generate the same topology. Thus, the metric spaces $\langle X,d_0\rangle$ and $\langle X,d_1\rangle$ share all purely topological properties, because they are identical as topological spaces: they are both $\langle X,\tau\rangle$, where $\tau=\tau_0=\tau_1$. Thus, they have the same open sets, closed sets, compact sets, connected sets, discrete sets, etc.
They do not necessarily share specifically metric properties. For example, one of $\langle X,d_0\rangle$ and $\langle X,d_1\rangle$ may be complete and the other not. One may be bounded and the other not. One may be totally bounded and the other not. They may differ in anything that depends specifically on the metric, rather than on the topology.