Proof. (attempt)
We verify that $\mathscr{B}$ is indeed a basis for a topology. Let $x \in X$. If $x$ is the smallest element, then $x \in [x,b) \in \mathscr{B}$. If $x$ is the largest element, then $x \in (a,x] \in \mathscr{B}$. Finally, if $x$ is neither then $x \in (a_{min}, b_{max}) \in \mathscr{B}$
Next, we need to prove that if $x \in B_1 \cap B_2$ where $B_i$ are basis elements, then can find some $B_3 \in \mathscr{B}$ so that $x \in B_3 \subset B_1 \cap B_2$.
If there are no smallest and no largest element, consider $B_1 = (a,b)$ and $B_2 = (c,d)$. If $B_1 \cap B_2 = \varnothing$, then we take $B_3 = \varnothing$. Otherwise $(a,b) \cap (c,d) = (c,b)$ so choose $B_3 = (c,b)$
Next, what if $X$ has largest element, then we need consider $(a,b) \cap (a,b_0]$ and if $X$ has smallest element, we need check $[a_0,b) \cap (a,b)$ in both cases, the intersection still gives elements in the basis.
IS this a correct proof?
The fast way is to note that if we define $$\mathcal{S} = \{(x, \rightarrow), (\leftarrow, x)\mid x \in X\}$$
then $\mathcal{B}$ is exactly the base generated by the subbase $\mathcal{S}$.
But if we follow the case distinction approach, we could refine your approach a litte bit and makeit more correct:
To see $\bigcup \mathcal{B} =X$.: let $x \in X$. If $x=\min(X)$, then $x \in [x,y) \in \mathcal{B}$ for any $y \neq x$ (using that $X$ has at least two points). If $x=\max(X)$, we use $x \in (y,x] \in \mathcal{B}$ in a similar way. If $x$ is not equal to $\min(X)$ or $\max(X)$ this means by definition that there are $y < x$ and $z> x$ in $X$, and then we can use $x \in (y,z) \mathcal{B}$.
Let $x \in B_1 \cap B_2$, where $B_1,B_2 \in \mathcal{B}$. If $x= \min(X)$ it follows that $B_1 = [x,a)$ for some $a \in X$ and $B_2 = [x,b)$ for some $b \in X$, as these are the only members of$\mathcal{B}$ that $x$ can be in if it is $\min(X)$. But then $x \in B_3:=[x,\min(a,b)) \subseteq B_1 \cap B_2$. The case that $x=\max(X)$ is similar. So we can assume $x$ is not $\min(X)$ nor $\max(X)$ and $B_1 = (a,b), B_2 =(c,d)$ for some $a,b,c,d \in X$ (if some $B_i$ happens to be of the special form $[\min(X),a)$, when $\min9X)$ exists e.g., we can replace it by $(\min(X),a)$ instead which is smaller and also contains $x$ in this subcase, and so if we have $B_3$ for that, then also for the original $B_i$ etc.) and such that $x \in (a,b)$ and $x \in (c,d)$. But then $B_3 := (\max(a,c), \min(b,d)) \in \mathcal{B}$ also contain $x$ and sits inside $B_1 \cap B_2$, as required.
So with explicitly checking all the cases it's a bit more involved, but not really hard, I'd say.