Let $\operatorname{Add}: \mathbb R^2 \rightarrow R$ and $\operatorname{Add}(x,y) = x + y$. Prove that is continuous using open sets.
Proof: $V \subset R$.
$$(\operatorname{Add})^{-1}(V) = \{ (x,y) \in \mathbb R^2: \operatorname{Add}(x,y) \in V\} = \{(x,y) \in \mathbb R^2 : x + y \in V\}$$
How to prove that $(\operatorname{Add})^{-1}(V)$ is an open set?
Let $V\subset{\mathbb R}$ be open, and assume $(x,y)\in{\rm add}^{-1}(V)$. We have to prove that there is a two-dimensional neighborhood $U$ of $(x,y)$ such that ${\rm add}(U)\subset V$.
Let $z:={\rm add}(x,y)\in V$. As $V$ is open there is an $\epsilon>0$ such that $\ ]z-2\epsilon,z+2\epsilon[\ \subset V$. The square $$U:=\ ]x-\epsilon,x+\epsilon[\ \times\ ]y-\epsilon,y+\epsilon[\ $$ is an open neighborhood of $(x,y)$. Choose an arbitrary point $(x',y')\in U$. A repeated application of the triangle inequality then shows that $$\bigl|{\rm add}(x',y')-z\bigr|\leq|x'-x|+|y'-y|<2\epsilon\ ,$$ hence $${\rm add}(x',y')\in V\ .$$