Can we give a metric $d$ on $\mathbb R$ such that the closure of $(0,1)$ with respect to $d$ is $[0,1]\cup\{2\}$?
I think the answer is no. For $x\in(0,1)$, there are disjoint open sets $U_x$ and $U_2$ (in the metric $d$) containing $x$ and $2$ respectively. That is for all $x\in(0,1),d(x,2)\gt0$ and $\{2\}$ is closed, so $\overline{(0,1)}\ne[0,1]\cup\{2\}$ in any metric $d$.
Is my argument correct?
I think one can even do it explicitly. Take e.g. the singular cubic curve parametrized by $$t \mapsto \big(t^2-1, t(t^2-1)\big), $$ which crosses itself at $t=\pm 1$, but is otherwise injective: if you reparametrize it so that this happens at $t=1,2$ (e.g. linearly by $t=2s-3$) you get the curve that is needed for your example. At this point, you can take $d$ to be the standard Euclidean metric in $\mathbb R^2$ and you can see that $(0,1)$ has limit points $\{0,1,2\}$ according to $d$.
Addendum: it is a nice exercise to show that $d$ actually satisfies the triangle inequality!
This is definitely possible, since a general metric on a set without any additional structure can be done in many ways. A metric on $\mathbb{R}$ is just a metrisable topology on a set of cardinality $2^{\aleph_0}$, and your question is asking if given such a set, a subset of the same cardinality could have exactly $3$ limit points added when taking its closure.
This is achieved by $[1,2) \cup (3,4) \cup (5,\infty)$. The closure of $[1,2) \cup (3,4) = [1,2] \cup [3,4]$ so we added exactly three points.
