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How to prove a commutative, with unit, Noetherian ring $A$ only has finitely many minimal prime ideals via the following step?
I have proved:
Step. All radical ideals of Noetherian ring $A$ can be expressed as an intersection of finitely many prime ideals.
How to continue?
If you know that radical ideals are finite intersections of primes, in particular $\sqrt{(0)}=\mathfrak p_1\cap\cdots\cap\mathfrak p_n$. Let $\mathfrak p$ be a minimal prime. Since $(0)\subseteq\mathfrak p$ we get $\sqrt{(0)}\subseteq\sqrt{\mathfrak p}=\mathfrak p$, that is, $\mathfrak p_1\cap\cdots\cap\mathfrak p_n\subseteq\mathfrak p$. It follows that there exists $\mathfrak p_i\subseteq\mathfrak p$ and since $\mathfrak p$ is minimal we must have $\mathfrak p_i=\mathfrak p$. (In other words, the minimal primes of $A$ are among the primes $\mathfrak p_1,,\dots,\mathfrak p_n$.)
Not sure about the approach you're taking but here is how one usually proves this fact.
Exercise 1: Show that in a ring $A$, the irreducible components of $\operatorname{Spec} A$ are in bijection with the minimal primes $\mathfrak{p}$. The bijection is given by $\mathfrak{p} \mapsto V(\mathfrak{p})$.
Exercise 2: Show if $A$ is Noetherian that $\operatorname{Spec} A$ is Noetherian. That is, it satisfies the DCC on closed subsets.
Exercise 3: Show that $\operatorname{Spec} A$ is the finite union of its irreducible components. (Hint: use the fact that every non-empty collection of closed sets ordered by inclusion in a Noetherian space has a minimal element).
Alternative approach: In a Noetherian ring $A$, $\text{Ass}(A)$ is a finite set. In a Noetherian ring every minimal prime is associated from which it follows that the set of minimal primes is finite too.
