So the lower limit topology on R has open sets of the form [a, b), where a<b, so far so good, my question is, if the arbitrary union of left closed sets can be open, that is for example $$\cup_\lambda^{\infty} [1/\lambda, 1)= (0,1)$$, where $\lambda\in N$, how does it fullfil the condition that the arbitrary unions of set in the collection is another set in the collection?
PS: I know this may be a dumb question, but the professor said it was trivial and I can't find a proof for the lower limit topology being a topology.
You are 100% correct. The collection of subsets of the form $[a,b)$ is not closed under arbitrary unions. You've observed correctly that $\bigcup_{n=1}^\infty [\frac{1}{n},1)=(0,1)$ which clearly is not of the form $[a,b)$. Another simple example is $\bigcup_{n=1}^\infty[0,n)=[0,\infty)$ which again is not of the form $[a,b)$. Another example is $[0,1)\cup[2,3)$ which is not of the form $[a,b)$.
So the collection of all subsets of the form $[a,b)$ is not a topology on $\mathbb{R}$. But that's not how the lower limit topology is defined. In fact it is defined to be the topology generated by all subsets of the form $[a,b)$.
So yeah, it is trivial. But nothing is trivial before it becomes trivial. Don't you love when a professor says "it is trivial" without actually explaining anything? :)
Note that the observation you've made has an interesting generalization. It can be shown that every interval of the form $(a,b)$ is open in the lower limit topology. Or equivalently that the lower limit topology contains the standard Euclidean topology.