Can one argue homeomorphism via equivalence of metrics? Let $S^n$ be the n-sphere with respect to $d_2$ metric (standart metric). Let $C^n$ be the n-sphere with respect to $d_1$ metric. Clearly we have,
$S^n$ $\cong$ $C^n$ (homeomorphism).
Yet, can one argue this is true because $d_1$ and $d_2$ are equivalent metrics? Would this idea generalize? 
 A: It is not a consequence of the equivalence of metrics. The reason it that the metrics $d_i$ on $\mathbb R^{n+1}$ are induced by norms $\lVert - \rVert_i$ on $\mathbb R^{n+1}$. It is well-known that all norms on $\mathbb R^{n+1}$ are equivalent, i.e. generate the same topology (the Euclidean topology) on $\mathbb R^{n+1}$. Hence each norm $\lVert - \rVert : \mathbb R^{n+1} \to \mathbb R$ is a continuous function with respect to this topology, and that is all we need to know.
So let us consider arbitrary norms $\lVert - \rVert_i$, i.e. $\lVert - \rVert_2$ is not necessarily the Euclidean norm $\sqrt{\sum_{i=1}^{n+1} x_i^2}$ and $\lVert - \rVert_1$ is not necessarily the  norm $\sum_{i=1}^{n+1} \lvert x_i \rvert$. Let $C_i = \{ x \in \mathbb R^{n+1} \mid \lVert x \rVert_i = 1 \}$ be the unit sphere with respect to $\lVert - \rVert_i$.
Define $h_1 : C_1 \to C_2, h_1(x) = x/\lVert x \rVert_2$ and $h_2 : C_2 \to C_1, h_2(x) = x/\lVert x \rVert_1$. Then
$$h_2(h_1(x)) = h_2(x/\lVert x \rVert_2) = \dfrac{x/\lVert x \rVert_2}{\lVert x/\lVert x \rVert_2 \rVert_1} = \dfrac{x/\lVert x \rVert_2}{(1/\lVert x \rVert_2) \lVert x \rVert_1} = x / \lVert x \rVert_1  = x$$
since $\lVert x \rVert_1 = 1$ for $x \in C_1$. Similarly $h_1 \circ h_2 = id$.
To see that the equivalence of metrics is not enough, consider the metric $d_2$ which gives you $S^n$. Define $d'_2(x,y) = \min(d_2(x,y), 1)$. This is an equivalent metric, but $\{ x \in \mathbb R^{n+1} \mid d'_2(0,x) = 1 \} = \{ x \in \mathbb R^{n+1} \mid \lVert x \rVert_2 \ge 1 \}$ which is not homeomorphic to $S^n$.
