As others commented there is no good way to metrize $Spec(A)$ as a topological space in general, because it is usually not Hausdorff. And we know that a serious metric space has to be Hausdorff and second countable. So you cannot use any kind of approximation arguments like we do in metric topology. However, for a lot of rings the Zariski topology is Noetherian and quasi-compact. So even though you cannot work with sequences, you can work open sets by open sets and using it not too difficult by principles like Noetherian induction. You can think of this as breaking up the whole space and work piece by piece to test the property you desired.
For your questions, let $A$ be a ring and let $Spec(A)$ be the set of prime ideals of $A$, with standard Zariski topology such that a basis for closed sets is given by $V(p), p\in Spec(A)$. Then the closure of a subset $O$ of $Spec(A)$ is given by the intersection of all closed sets that contains $O$. You may write it as follows:
$$
\overline{O}=\bigcap J, J \textrm{closed in Spec A}, O\subset J
$$
The claim is $\overline{O}=V(I(O))$, where $I$ is the intersection of all prime ideals containing (prime ideals of $O$). This is one side of the so called closed subset - radical ideal correspondence when $A=k[x_{1}\cdots x_{n}]$, the so called affine coordinate ring.
Now if a prime ideal containing $I(O)$, then it must contain at least one of the prime ideals in $O$. And inducting forwards we can show $V(I(O))=\bigcup_{p\in O}V(p)$. To see this let $I(O)=\bigcap_{I}p_{i}$ and assume $p\supset \bigcap_{i\in I'}p_{i}\supset \bigcap_{i\in I}p_{i}$, where $I'$ is finite. If $p$ does not contain any of the $p_{i}$s, then we can choose elements from $p^{c}\cap p_{i}$ for all $i\in I'$, then multiply them together. The product will be an element in $p$, while none of the factors is in $p$, which contradicts the fact that $p$ is prime. Therefore $p$ must contain at least one $p_{i}$ in $O$. We now claim that $p$ must be contained in all closed subset $J$ that contains $O$.
To see this, let $q$ to be an element not in $\overline{O}$. Then there exists an open set $O_{q}\cap O=\emptyset$. The complement of $O_{q}^{c}$ is given by a set of type $V(E)$ for some $E\subset A$. Since $p_{i}\in O_{q}^{c}=V(E)$, $p_{i}\supset E$ and $q\in D(E)$ does not contain $E$. Therefore $q$ does not contain $p_{i}$ as well. In particular $p$ cannot be $q$. Therefore $p$ must be in $\overline{O}$. This showed that $V(I(O))\subset \overline{O}$. The other direction $\overline{O}\subset V(I(O))$ should be immediate. Therefore $\overline{O}=V(I(O))$ as desired. For the case $p$ is a point, the proof can be simplified as we do not need the first part, and the second part should still carry through.
This "proof" by me might be unnecessarily long, and I welcome anyone who can provide a slick shorter proof. But I hope I have covered all detail. There is a gap that it is possible that $p\not \supset \bigcap_{I'}p_{i}, i\in I'$, for all $I'$ finite while $p$ covers $\bigcap_{I}p_{i},i\in I$. If this is the case, obviously $p\not \supset p_{i}$ for all $i$. And here I got more or less stuck. This has been constructed in an explicit example at mathoverflow.
