Show that any rectangle $(a, b)\times (c, d)$ is open in $\mathbb R^2$. I can easily prove this for Euclidean metric.  But here the metric is not specified.
So I need a generalized proof for all metrics.
How do we start then?
 A: One way to give a metric topology on the plane in which not all rectangles $(a,b)\times (c,d)$ are open is by imposing a topology that makes the plane homeomorphic to the real line.  In greater detail:
We know that in Euclidean topologies $\mathbb{R}$ is not homeomorphic to $\mathbb{R}^2$.  For example, removing a point separates the real line, but does not separate the plane.
However we do have $\mathbb{R}$ and $\mathbb{R}^2$ of equal cardinality, so let $f:\mathbb{R} \to \mathbb{R}^2$ be a one-to-one correspondence.   Let $\tau$ be the metric topology on $\mathbb{R}^2$ induced by this correspondence with the usual metric on $\mathbb{R}$.  Thus $f$ is continuous with respect to the topology $\tau$ on the plane.
If all rectangles $(a,b)\times (c,d)$ in the plane were open in the metric topology $\tau$, then the identity map $id:\mathbb{R}^2 \to \mathbb{R}^2$ with respect to the topology $\tau$ on the domain and the usual topology on the codomain would be continuous and injective.
It follows by invariance of domain  that the composition of maps $id(f)$ would be a homeomorphism of $\mathbb{R}$ and $\mathbb{R}^2$.  Contradiction.
A: Here is another example. Let $f : \mathbb{R}^2 \to \mathbb{R}^2$ be the bijection which fixes all points $p \in \mathbb{R}^2$ that are not equal to $(0,0)$ or $(2,0)$ but swaps those two points:
$$f(p) = \begin{cases}
p &\text{if $p \ne (0,0)$ and $p \ne (2,0)$} \\
(2,0) &\text{if $p = (0,0)$} \\
(0,0) &\text{if $p = (2,0)$}
\end{cases}
$$
Let $\mathcal{T}_1$ be the usual metric topology on Euclidean space, with metric denoted $d_1$.
Let $\mathcal{T}_2$ be the topology
$$\mathcal{T}_2 = \{f(U) \bigm| U \in \mathcal{T}_1\}
$$
which is also a metric topology: $d_2(p,q) = d_1(f^{-1}(p),f^{-1}(q))$.
The set $V=(-1,1) \times (-1,1)$ is open in $\mathcal{T}_1$. But it is not open in $\mathcal{T}_2$, because if $V=f(U)$ then
$$U = \bigl(V - \{(0,0)\}\bigr) \cup \{(2,0)\} \not\in \mathcal{T}_1
$$
