This is purely based on my understanding only.
Let's start with σ-algebra:
If X is a set, then all subsets of F together can be collected into the power set of X. Take any subset of the power set, say F (it is a set which is a collection of sets, or simply a collection). This obviously doesn't have all of the subsets of X. But it does have a few. If,
i)X is included in the collection F,
ii) all subsets in F have their complements also in F, and
iii) countable unions (even intersections!) of sets in F are also available in F,
then we call F a σ-algebra.
What if our collection is not a σ-algebra? Say our collection of subsets of X is S. We can still add a few subsets of X to S so that we get a σ-algebra. The smallest such σ-algebra is the σ-algebra generated by S.
Now, essentially, a σ-algebra and a topology are quite different. While σ-algebra is closed under countable unions and complements, a topology is closed under arbitrary (finite or infinite) unions and finite intersections.
{Unnecessary info: So, σ-algebra imposes the additional condition of closure under complements on topology. (I am not quite sure about this!)}
An open set can be an element of a topology. A set is called "open" if, for every point inside that set, you can always find a smaller open set around that point that is completely contained within the original set. You can zoom in a bit for every point and stay entirely within the set. Open sets can come together to define a topological space satisfying all its properties.
Now, [0,1] in your example is not an open set. This is because, for the points 0 or 1 in [0, 1], you can't find another open set entirely within [0, 1] that contains 0 or 1.
The Borel σ-algebra is defined as the σ-algebra generated by the open sets. Take the set (0,1]. Consider any collection of subsets of (0,1]. Take {(0,0.5)} for example. This is a singleton collection of an open set. To generate a σ-algebra, add ∅,[0,1]. Add complement of (0,0.5), i.e., [0.5,1]. Borel σ-algebra generated by {(0,0.5)} is hence {∅, (0,1], (0,0.5), [0.5,1]}. The sets ∅, (0,1], (0,0.5) and [0.5,1] are all Borel sets. Can {[0.5,1]} generate the same? Yes! As a consequence of the condition of closure under complements, the closed set [0.5,1] also generates the same.
So your example [0,1], which is a closed set, also generates a Borel σ-algebra. Any open interval subset of [0,1] is a Borel set.
The class B of Borel sets in Euclidean space is the smallest collection of sets that includes the open and closed sets such that the countable union of sets in B are also included in B.
Correct me if I am wrong!