The short answer to "Is there any connection between homeomorphism and equivalence of metric spaces?" is yes. The long answer: any reasonable notion of equivalence of two metrics $d_1$ and $d_2$ can be formulated in terms of the identity map $\mathrm{id}\colon (X,d_1)\to (X,d_2)$. As soon as we distinguish a class of "nice" maps (the class should be a group under composition), we get a notion of equivalence. Some examples, previously mentioned and not:
$\mathrm{id}$ is a homeomorphism
$\mathrm{id}$ is a uniform homeomorphism (i.e., uniformly continuous with a uniformly continuous inverse)
$\mathrm{id}$ is a bilipschitz homeomorphism (i.e., Lipschitz with a Lipschitz inverse)
$\mathrm{id}$ is a quasisymmetric homeomorphism
$\mathrm{id}$ is an isometry
$\mathrm{id}$ is a quasi-isometry
... The list is not exhaustive.
The corresponding notions of equivalence are related by $5\implies 3\implies 2\implies 1$, also $3\implies 4\implies 1$, and $3\implies 6$.
Such metrics are also called Lipschitz equivalent, the inequality states that the identity map from the space endowed with one metric to the space endowed with the other is a Lipschitz map (in both directions). That ensures that the space is complete in one metric if and only if it is complete in the other metric.
In fact, that holds for a weaker concept of equivalence, uniform equivalence of metrics. Two metrics $d_1,d_2$ on a space $X$ are uniformly equivalent if the identity map $(X,d_i) \to (X,d_j)$ is uniformly continuous for both choices of $\{i,j\} = \{1,2\}$. With formulae, the metrics are uniformly equivalent if for every $\varepsilon > 0$ there are $\delta_1, \delta_2 > 0$ such that
$$\bigl(d_1(x,y) \leqslant \delta_1 \implies d_2(x,y) \leqslant \varepsilon\bigr) \land \bigl(d_2(x,y) \leqslant \delta_2 \implies d_1(x,y) \leqslant \varepsilon\bigr).$$
Since uniformly continuous maps - in particular Lipschitz continuous maps - map Cauchy sequences to Cauchy sequences, uniform equivalence of the metrics implies that a sequence is a $d_1$-Cauchy sequence if and only if it is a $d_2$-Cauchy sequence. Since uniformly equivalent metrics (and hence in particular Lipschitz equivalent metrics) induce the same topology (i.e. they are topologically equivalent), a sequence is $d_1$-convergent if and only if it is $d_2$-convergent.
Hence, if the metrics are uniformly equivalent, the space is complete in one metric if and only if it is complete in the other. And by contrapositive, if you have two metrics on a space such that the space is complete in one but not in the other, the two metrics aren't uniformly equivalent, and a fortiori not strongly equivalent.
Best Answer
Work in $\mathbb R^2$ since the idea carries over easily to higher dimensions. Let $(x,y) \in \mathbb R^2$ and assume (with no loss of generality) that $|x| \le |y|$.
Since $|x|^2 + |y|^2 \le 2|y|^2$ you have $$\sqrt{x^2 + y^2} \le \sqrt 2 |y| = \sqrt 2 \max \{|x|,|y|\}.$$
Since $|x| \le |y|$ you have $$\max\{|x|,|y|\} = |y| = \sqrt{y^2} \le \sqrt{x^2 + y^2}.$$