I am trying to solve this problem.
Let $I$ be a non trivial ideal of a ring $A$, and let $r(I)$ be its radical. Prove that $I=r(I)$ iff $I$ is an intersection of prime ideals.
I got the right to left implication:
If $I$ is an intersection of prime elements, let $x\in I$. Since $x^1\in I$, we know $x\in r(I)$, and thus, $I\subseteq r(I)$. On the other hand, let $x\notin I$ such that $x^n\in I$. Then, since $I$ is an intersection of prime ideals (let them be $P_1, ..., P_k$), $x^n\in P_j$, for all $j$, and since all $P_j$ are prime, either $x\in P_j$, or $x^{n-1}\in P_j$, for all $j$. Repeating this reasoning, we end up with $x\in P_j$, for all $j$, and thus $x\in I$, which is a contradiction. Therefore, if $x\in r(I)$, then $x\in I$, and $I=r(I)$.
On the left to right, I'm struggling a little. I'm trying to prove that $I$ is a prime ideal, and therefore I'll get $I$ as an intersection of a single prime ideal, but maybe that's not the way to go.
Also, is my reasoning correct?
Slight correction first: $r(I)$ being an intersection of prime ideals does not mean that $r(I)=P_1\cap\dots\cap P_k$ for prime ideals $P_1,\dots,P_k$, but rather $r(I)=\bigcap_{i\in I}P_i$ for prime ideals $P_i$ and some index set $I$. However, your argument for the backwards direction remains intact.
As already pointed out in the comments, the notion of radical and prime ideals are distinct. Moreover, the standard approach to this statement is more naturally formulated as the following fact.
Clearly, $r(I)$ is contained in the given union. Indeed, if $x\in r(I)$, then $x^n\in I$ for some $n>0$. Hence, inductively, $x\in P$ for any prime ideal $P$ containing $I$. This is essentially the argument you gave.
The converse is a bit more tricky. I will follow the proof idea of Atiyah--MacDonald, as mentioned in the comments.
We show the contrapositive, that is, if $x^n\notin I$ for any $n>0$, then there is a prime ideal (in fact, a maximal ideal) $P$ such that $x\notin P$ for all $I\subseteq P$. This is essentially done by considering the localisation of $A$ at $x$, i.e. the set of formal fractions $a/x^n$, $n\ge0$. However, we can also spell out the main incredient of the above reudction, which is an application of Zorn's Lemma. Consider the set $$ \Sigma =\{J\,|\,I\subseteq J,\,x^n\notin J\text{ for all }n>0\}\,. $$ This set is non-empty, since $I\in\Sigma$ by assumption, and the restricting condition is stable under unions along chains. Thus, every chain has an upper bound, hence $\Sigma$ admits a maximal element. This maximal element, $P$, is prime. Indeed, if $a,b\notin P$, then $P+(a)$ and $P+(b)$ properly contain $P$, hence $P+(a),P+(b)\notin\Sigma$. Hence, we find $n,m>0$ such that $x^n\in P+(a)$ and $x^m\in P+(b)$. Then $x^{n+m}\in P+(ab)$ and therefore $ab\notin P$, as otherwise $P+(ab)\in\Sigma$.