Union of bases on $X$ is not a base. Let $X$ be some set, and $B_1,B_2$ be some bases on $X$ that generate some topologies. I want to double check that it's not true in general that $B_1\cup B_2$ is going to be a base on $X$.
Indeed, assume that $B_1\cup B_2$ is a base on $X$. consider $X=\mathbb{R}$ and take
$$B_1=\{(a,b]|a,b\in\mathbb{R}\},$$
$$B_2=\{[c,d)|c,d\in\mathbb{R}\}.$$
Then we can see that one of the property for the base $B_1\cup B_2$ is not satisfied as
$$(0,1]\cap[1,2)=\{1\}.$$
So, we cannot find elements of $B_1$ or $B_2$ such that they are contained in the intersection $(0,1]\cap[1,2)$. Therefore, $B_1\cup B_2$ is a base on $X$ as we found a counterexample.
Also, I have a follow up question. Assume that we have two topologies $\tau_1$ and $\tau_2$ on $X$. I want to show that there exists a unique smallest topology $\tau$ on $X$ that contains $\tau_1$ and $\tau_2$. Is it enough to take $\tau_1\cup\tau_2$ and create $\tau$ by closing $\tau_1\cup \tau_2$ under all possible unions and intersections?
 A: As far as bases are concerned, you are absolutely right, your counterexample works just fine.
I would just remove one thing from your argument. The sentence "assume that $B_1\cup B_2$ is a base on $X$" is completely redundant, and it obfuscates, rather than clarifies your proof. It makes it sound as if your proof is a proof by contradiction, when in fact, it is not. Your proof is a proof by counterexample, and that proof does not need the "assume $X$, therefore blahblah, therefore not $X$) structure.
A proof by counterexample has a simpler structure, wherein you prove $\neg(\forall x: P(x))$ by simply proving $\exists x: \neg P(x)$, and that is exactly what you did.

For the topologies, you could do what you say, or you could notice that the intersection of any collectioni of topologies is also a topology, so you could just take the intersection of all topologies that include $\tau_1\cup\tau_2$, and notice that this intersection is indeed the smallest topology that includes both $\tau_1$ and $\tau_2$.
