This question was asked in a master's exam for which I am preparing and I was unable to find out which result I should use. So, I am posting it here.
Question: (i) does there exists a metric on which $(0,1)$ is complete? (More generally can you please tell if there always exits a metric such that the space which is fixed while finding the metric is complete?)
(ii) Does there exists a metric such that $[0,1]$ is not complete?
I studied analysis from Apostol but unable to think on how should I approach this particular question (i.e. which result I should) use. There were 2 more options in question but I worked them out.
Suppose you have a metric space $(X, d)$, a set $Y$ (no topological features necessary), and a bijection $\varphi : Y \to X$. Then $$d' : Y \times Y \to \Bbb{R} : (y_1, y_2) \mapsto d(\varphi(y_1), \varphi(y_2))$$ defines a metric such that $\varphi$ becomes an isometry between $(X, d)$ and $(Y, d')$. In particular, $(Y, d')$ is complete if and only if $(X, d)$ is complete.
In this case, we can put $(0, 1)$ in bijection with $[0, 1]$, both having cardinality of the continuum. If we inherit the metric from $[0, 1]$, then $(0, 1)$ becomes complete. If we inherit the metric from $(0, 1)$, then $[0, 1]$ becomes incomplete.
EDIT: You can also make $(0, 1)$ complete by putting it into bijection with $\Bbb{R}$ using a homeomorphism like $$\varphi : (0, 1) \to \Bbb{R} : x \mapsto \tan\left(\pi\left(x - \frac{1}{2}\right)\right).$$ This shows that you can preserve the topology of $(0, 1)$, but yet change whether the space is complete.
On the other hand, we cannot form a homeomorphism between $[0, 1]$ and an incomplete space, as the continuous image of $[0, 1]$ is necessarily compact, and hence complete.