Problem on completeness. Which of the following is/are true? 


*

*(0,1) with usual topology admits a metric which is complete. 



*(0,1) with usual topology admits a metric which is not complete. 

*[0,1] with usual topology admits a metric which is not complete. 

*[0,1] with usual topology admits a metric which is complete.



Since every subspace of a metric space is complete iff it is closed. 
So R taken as topological space with usual topology and (0,1) &[0,1] as subspace of R .then (0,1) is open and [0,1] is closed. 
So options should be true 2&4.
But options are true 1,2&4.(According to CSIR) 
Also I'm seeing 1&2 are negation of each other. 
Where is my misunderstanding? Please convince me!! I will appreciate any efforts by you. Thanks in advance. 

$3$ is false due to $[0,1]$ being compact in the usual topology (see Every compact metric space is complete), and $2$ and $4$ are true for the reason you mention (with the metric inherited from $\mathbb R$).
Now to $1$: We could imbue $(0,1)$ with a new metric, not inherited from $\mathbb R$, in which it will be complete, yet which will produce the same topology as usual. Such a metric space will be a subspace of $\mathbb R$ as a topological space but not as a metric space. (Thus the theorem you cited doesn't apply.)
One such example would be $d(x,y)=|f(x)-f(y)|$ where $f(x)=\tan\left(\pi\left(x-\frac{1}{2}\right)\right)$. That would make $(0,1)$ isometric with $\mathbb R$ in the usual sense, and therefore it will be complete.
(Note: $f$ could have been chosen as any homeomorphism of $(0,1)$ to $\mathbb R$, the 'tangent' function is merely a convenience.)
