Recall the definition. A topological space is a pair $(X,\mathcal{T})$where $X$ is a set and $\mathcal{T}$ (is defined to be the set of open sets,) $\mathcal{T}$ is a collection of subsets of $X$ such that
- $\emptyset\in\mathcal{T}$
- Any union $A$ of a collection of $A_i\in\mathcal{T}$ is also in $\mathcal{T}$
- Any finite intersection $A$ of $A_1,\ldots,A_n\in\mathcal{T}$ is also in $\mathcal{T}$.
Then the topology generated by a set $\mathcal{A}$ is the coarsest ("smallest") topology $\mathcal{T}$ such that $\mathcal{A}\subseteq\mathcal{T}$.
So an open space in the topology genrated by the intervals $B=\{[a,b)\subseteq\Bbb{R}:a<b\}$ has the open sets $A\in\mathcal{T}$ such that $A$ can be obtained via taking arbitrary unions or finite intersections of $B$-intervals. For example, $[0,2)$ is open, as is $\bigcup_{n\ge 1}[0,n)=[0,\infty)$. For a more interesting example, consider the set given by
$$\bigcup_{n\ge 2}\,[1/n,1),$$
which is also open, as an arbitrary union of open sets; notice this union is equal to $(0,1)$.
On a side note, $(0,1]$ is in fact not open. To show a set is not open, one way is to show that the complement is not closed. In your example, to show that $(0,1]$ is not open in the topology generated by $B$, I would show that $(-\infty,0]\cup(1,\infty)$ is not closed in the topology generated by $B$.
Lemma. A subset $C$ is closed iff every convergent (in $X$) sequence $x_n\to x$ with points $x_n\in C$ has $x\in C$.
To show that $x_n\to x$ in the topology generated by $B$ it is sufficient to show that for every set in the basis $[a,b)\in B$, that contains $1$, there is some $N$ such that $n\ge N$ implies $x_n\in [a,b)$. Let $x_n=1+1/n$, I claim that $x_n\to 1$ in $(\Bbb{R},B)$. For every base element containing $1$ (i.e. $[1,1+\varepsilon)$), there is an $N$ such that $n\ge N$ implies $1/n<\varepsilon$, i.e. $x_n\in [1,1+\varepsilon)$; but $1\notin(-\infty,0]\cup(1,\infty)$. Hence, $(-\infty,0]\cup(1,\infty)$ is not closed, so $(0,1]$ is not open.
Best Answer
The set is actually neither open nor closed. (Perhaps these words are a bad choice cause they seem to imply mutual exclusiveness when it is not the case. Sets may be open, closed, both, or neither.)
It is not open for the same reasons you said in your question.
A set is closed if and only if its complement is open. The endpoints of the interval are in the complement of the interval and you can't make a ball around them without including points that are not in the complement. Thus the complement is not open so the interval is not a closed set.
Openness and closedness always depend on the total space. For instance any subset $S\subset \mathbb R^n$ is 'open' when you consider the $S$ as the topological space (and give it the subspace topology) as it must be to satisfy the axiom of topology that says the total space is open.