$[a,b)$ not closed I understand this can easily be done by contradiction by I am slightly confused if I use a different method.
If $(x_n)$ is an arbitrary sequence, $x_n \in K$,  $\forall n>0$ and $\lim \limits_{n \to \infty}x_n=x$ with $x \in \mathbb{R}$ then $x \in K \iff$ $K$ is closed
The negation of this is:
If $\exists$ a sequence $ (x_i)$  with $x_i \in K$ (where $K$ is a  set) $\forall i>0$ and $\lim \limits_{i \to \infty}x_i=x$ with $x \in \mathbb{R}$ then $x \notin K \iff$ $K$ is not closed
However my lecturer has used $x_i=b-\frac{1}{i}$ which has the limit $b$ which is what we want as $b \notin [a,b)$, however $x_i$ does not necessarily $\in [a,b), \forall i>0$? For example $x_1=b-1$ may well be less than $a$ and hence $\notin [a,b)$?
 A: You have a great point.  But still, there is no problem here.  This is because with sequences, we are almost always concerned with the end behavior of the sequence.
If $x_{i}$ is not in $[a,b)$ for the first few $i$, that's okay.  We could just define a new sequence $y_{i}$ to be the sequence $x_{i}$ but only at the starting point where all successive terms are in $[a,b)$.  I hope you understand why if $x_{i} = b - \frac{1}{i}$, after a certain finite point, all successive terms will be in $[a,b)$.  And so, we could just as easily ignore all of those finitely many terms that are outside of $[a,b)$ and just consider the sequence gotten by starting at the first point of $x_{n}$ where all successive terms are in $[a,b)$.
A:   (b-(1/n))converges to b. So, one can find m in IN such that b-(1/n) is in (a,b+1) for n>=m. 
  => b-(1/n) is in (a,b) for n>=m.(for each such term is < b)

   Also, (b-(1/m),b-(1/(m+1)),...) converges to b. Take x_1 = b-(1/m), x_2 = b-(1/(m+1)),.... Maybe, your lecturer wanted you to work with some tail of (b-(1/i)).

