Proof that $[a,b)$ is neither closed nor open? Our professor gave us definitions for closed and open intervals. 
A set $U$ is open if $\forall x \in U$, $\exists \epsilon \gt 0$ such that $(x- \epsilon,x+ \epsilon)\subset U$. A set is $F$ closed, if its complement is open.
I'm not too sure how to write the solution that is clear and understandable. This is what I've thought of so far. 
To prove that the set $U=[a,b)$ is not open, I used proof by contradiction by assuming that it is open. Since $a \in U$, it follows that $\exists \epsilon \gt 0$ such that $(a- \epsilon,a+ \epsilon)\subset U$. However, this is not possible as there does not exist $\epsilon \gt 0$ such that $a- \epsilon \in U$. The proof that the set is not closed is similar to this.
 A: Welcome to MSE!
$U=[a,b)$ is not open means $$(\color{red}{\exists x} \in U \wedge\color{blue}{\forall \varepsilon>0}):(x-\varepsilon,x+\varepsilon) \subset U\;\;\text{is false}$$
You are right about that $x$ namely $x=a$. You assume the left endpoint of the nbd of $a$ does't in $U$. But $ a- \varepsilon$ is  not even in $(a-\varepsilon,a+\varepsilon)$. Note that  every element of the portion $(a-\varepsilon,a)$  is not  in $U$, so $U$ is not open 

For other $x$'s in $U$,  we have $$(x-\varepsilon,x+\varepsilon) \subset U$$ How to choose $\varepsilon$ in this case?

A: Seems correct.
If the set is closed, then $ I^{\text C}:=]-\infty,a[\cup[b,\infty[ $ is open; since $ b\in I^{\text C} $, $ ]b-\epsilon,b+\epsilon[ $ should also be contained in $ I^{\text C} $, for some $ \epsilon > 0 $, which is a contradiction.
Of course you need to account what @Chinnapparaj R said, i.e. to show that there is $ \mu\in ]b-\epsilon,b[ $ that makes $ ]b-\epsilon,b[ \not\subset I^{\text C} $
A: You have a slight error.
You say there is no $\epsilon$ so that $a -\epsilon$ is in $U$.  But there is nothing in the definition that says $a -\epsilon \in U$.  Then definition says $(a-\epsilon, a + \epsilon) \not \subset U$.  ANd $a-\epsilon \not \in (a-\epsilon, a+\epsilon)$ so that does not contradict.
However as there exist for any $\epsilon > 0$ many (uncountably many) $y \in (a-\epsilon, a)$ so that $a-\epsilon < y < a$ and in particular, there is  $a  - \frac \epsilon 2 \in (a-\epsilon, a + \epsilon)$.  $a-\epsilon$ is NOT a valid counter example point.   But $a - \frac \epsilon 2$ (as is $a-.99999\epsilon$) is.  And from that your argument follows.
......
To prove in is not closed is similar.  
But let's go into detail.  $b \not \in [a,b)$ so $b \in [a,b)$.  ANd for $b$ there is no $\epsilon > 0$ so that $(b-\epsilon,b+\epsilon) \subset [a,b)^c$.  
Why not?  Because there are infinitely many $y$ so that $\max(a, b-\epsilon) < y < b$ and these are all in $y \in [a,b)$ so $y \not \in [a,b)^c$.
To not twist ourselfs into knots it is probably easier to put it this way:
If $[a,b)^c$ is open then there will be an $\epsilon > 0$ so that $(b-\epsilon, b+\epsilon)\subset [a,b)^c$.  We might as well assume $0 < \epsilon < b-a$.  [We can assume that because for any $\epsilon$ so that $(b-\epsilon, b+\epsilon ) \subset U$ then for any $0 < \epsilon_2 < \epsilon$ we'd have $(b-\epsilon_2,b+\epsilon_2) \subset (b-\epsilon, b+\epsilon ) \subset U$ ans we can always find an $0 < \epsilon_2 < b-a$]
Then let $a < b - \frac \epsilon 2 < b$ so $b-\frac \epsilon 2 \in [a,b)$ and $b-\frac \epsilon 2\not \in [a,b)^c$.  But $b-\frac \epsilon 2 \in (b-\epsilon, b+\epsilon)$ so $(b-\epsilon, b+\epsilon)\not \subset [a,b)^c$.  So $[a,b)$ is not closed.
....
Hmm, maybe that wasn't easier. 
