Proving that principal open sets form a base for the Zariski topology

The Question

I believe I have proven the following question, and would like feedback on my proof.

Specifically, I am uncertain on these points:

• How to approach the fact that $$A\cong k[x_1,\dots,x_n]/I$$ for an ideal $$I\subset k[x_1,\dots,x_n]$$, instead of the familiar case when $$A=k[x_1,\dots,x_n]$$.
• The difference between the Zariski topology on $$\mathbb{A}^n$$ versus on $$\operatorname{m-Spec}(A)$$.

Furthermore, I will bold sentences that I am unsure I am able to make.

Fix an algebraically closed field $$k$$. Consider the Zariski topology on $$\operatorname{m-Spec}(A)$$, where $$A$$ is a finitely generated, reduced $$k$$-algebra. Given $$f\in A$$, consider its closed zero set $$Z(f)\subseteq \operatorname{m-Spec}(A)$$ and its open complement $$U_f\subseteq\operatorname{m-Spec}(A)$$. These are called principal open sets.

Show that the principal open sets form a base for the Zariski topology.

The Proof

To show that these open sets form a base for the Zariski topology $$\mathcal{T}$$, we check the following condition:

• Given an open set $$U\in\mathcal{T}$$, if $$p\in U$$ then there is some $$f\in A$$ such that $$x\in U_f\subseteq U$$.

Let $$U\in\mathcal{T}$$ be an open set in the Zariski topology, and consider an element $$p\in U$$. This implies that for some ideal $$I\in\operatorname{m-Spec}(A)$$, there exists some $$f\in I$$ such that $$f(p)\ne 0$$.

Consider the principal open set $$U_f$$. To show that $$U_f\subseteq U$$, we note that if $$p'\in U_f$$, then $$f(p')\ne0$$, and since $$f\in I$$, it follows that $$p'\notin Z(I)$$ and so $$p'\in U$$. Hence, $$U_f\subseteq U$$, and so the principal open sets form a base of the Zariski topology.

Actually, the conditions you've mentioned only show that the $U_f$ form a base for some topology on $X$. You can't be sure a priori that this topology will be the Zariski topology (notice that you never even used any properties of the Zariski topology in your proof!).
If you already have a topology on a space $X$, and you want to show a collection of open sets $\mathcal B$ form a base for your specific topology, then you just need to show that given an open set $U\subseteq X$ and a point $x\in U$, there exists some $B\in\cal B$ such that $x\in B\subseteq U$.
Edit: your new proof is correct, though maybe you'd wanna write more explicitly in the first paragraph that the complement of $U$ is closed, so of the form $Z(I)$, and this is the $I$ you're using through the rest of the proof.