Let $\tau$ be the topology on $\mathbb{R}$ generated by $B = \{[a, b)\subset \mathbb{R}: -\infty<a<b<\infty\}$. Then the set $\{x\in \mathbb{R}: 4\sin^2x\leq 1\}\cup \big\{\frac{\pi}{2}\big\}$ is closed in $(\mathbb{R}, \tau)$.
My effort:
We know that $4\sin^2x\leq 1$ whenever $-\frac{1}{2}\leq \sin x \leq \frac{1}{2}$, i.e. $x \in \big[-\frac{\pi}{6}, \frac{\pi}{6}\big]\cup \big[-\frac{11\pi}{6}, \frac{13\pi}{6}\big]\cup\cdots$. How to proceed further?
On
One way to show that a set is closed in a topology is to show that its complement is open. In this particular topology, any open interval is open:
$$(a, b) = \cup_{n \in \Bbb N} [a-\frac{1}{n}, b).$$
And it's easy to see that your set is the complement of a union of open intervals, which we now know are open in $\tau$, so we're done.
Since $\left\{\frac\pi2\right\}$ is closed in $(\mathbb{R},\tau)$, all you need to do is to prove that$$\cdots\cup\left[-\frac\pi6,\frac\pi6\right]\cup\left[-\frac{11\pi}{6}, \frac{13\pi}{6}\right]\cup\cdots$$is closed there. But its complement is an union of intervals of the type $(a,b)$, and these intervals are open in $(\mathbb{R},\tau)$.