For an open interval $(a,b)$ in $\mathbb R$, show that $\left\{(x,y)\in R^2 : x\in (a,b)\wedge y\in\mathbb R\right\}$ is open in $\mathbb R^2$. I would prefer helpful hints and guidance over an explicit answer. I am having serious difficulty understanding how to show that a set is open, I feel as though it should be easy but right now it seems difficult.
Since $(a,b)$ is open I know for $x\in (a,b)$, there exists $(x-\epsilon, x+\epsilon)$ contained in $(a,b)$, and I understand I essentially have to show that for an arbitrary point in our set, there is an epsilon-ball about our point entirely contained in our set. from here, I feel overwhelmed and almost completely lost.
Note: this may not be exactly the answer you are looking for, but it is a point of view worth keeping in mind.
By definition, the product topology is the coarsest (that is, the one with the least possible open subsets) that satisfies the property that the projections on each factor $$\operatorname{pr}_1, \operatorname{pr}_2: \Bbb R\times \Bbb R\to \Bbb R$$ are continuous maps.
Now $\operatorname{pr}_1$ being continuous implies that the preimage $\operatorname{pr}_1^{-1}(U)$ is open in $\Bbb R\times \Bbb R$ for any open subset $U\subset \Bbb R$.
Applying this to $U=(a,b)$ gives you exactly what you want.