Let $A$ be a ring and $X=spec(A)$, the prime spectrum of $A$. Prove that $X$ is quasi-compact.
Definition of quasi compact: each open covering of $X$ has a finite subcovering of $X$.
It is enough to considering the covering in the basis $\{X_f|f\in A\}$.
Let the set of $\{X_{f_i}| f_i\in A , i\in I\}$. I is some index set. It is obvious that $f_i$ with $i\in I$ generates the unit ideal. But why there exists a finite subset $J$ of $I$ such that $f_i$ with $i\in J$ generates the unit ideal?
Hint. $\langle f_i\rangle_{i\in I}=\langle1\rangle$ if and only if $1\in\langle f_i\rangle_{i\in I}$. What does an arbitrary element of $\langle f_i\rangle_{i\in I}$ look like and use the latter statement I gave.