I'm going to provide a detailed proof for this theorem:
Let $G$ and $H$ be non-trivial simple groups. If there is a prime number $p$ with $$|G|=|H|=p$$ then $G\times H$ has $p+3$ normal subgroups. Otherwise $G\times H$ has $4$ normal subgroups.
Is there anything wrong with this proof:
Proof: Let $N$ be a normal subgroup of $G\times H$. As the projection mappings $\pi_1:G\times H\to G$ and $\pi_2:G\times H\to H$ are onto homomorphisms, we have $$\pi_1(N)\trianglelefteq G, ~~\pi_2(N)\trianglelefteq H $$ So one of the following holds:
1) $\pi_1(N)=\{1\},~ \pi_2(N)=\{1\}$: We have
$$(\forall (g,h)\in N)(g = 1, h= 1)$$ $$\Rightarrow N=\{(1,1)\}$$
2) $\pi_1(N)=G,~ \pi_2(N)=\{1\}$: We have
$$(\forall (g,h)\in N)( h= 1)$$ $$\Rightarrow N=\pi_1(N)\times \{1\}=G\times \{1\}$$
3) $\pi_1(N)=\{1\},~ \pi_2(N)=H$: We have
$$(\forall (g,h)\in N)(h= 1)$$ $$\Rightarrow N= \{1\}\times\pi_2(N)= \{1\}\times H$$
4) $\pi_1(N)=G,~ \pi_2(N)=H$: Let $a,b\in G$ be arbitrary. There is some $y\in H$ with $$(b,y)\in N$$ We have $$([a,b],1)=(aba^{-1}b^{-1},1)=(a,1)(b,y)(a,1)^{-1}(b,y)^{-1}\in NN^{-1}=N$$ So $$G'\times \{1\}\subseteq N$$ Similarly
$$\{1\}\times H'\subseteq N$$ So $$G'\times H'\trianglelefteq N$$ So if at least one of $H$ or $G$ is not abelian $N=G\times H$. Suppose both $H$ and $G$ are abelian. They are both simple so both are finite and have prime order: $$|G|=p, ~~ |H|=q$$ $G\times H$ is isomorphic to $\Bbb Z_p\times \Bbb Z_q$.
If $p\ne q$, $\Bbb Z_p\times \Bbb Z_q$ has only 4 subgroups. So $N$ is either $\{(1,1)\}$ or $\{1\} \times G$ or $G\times \{1\}$ or $G\times H$.
If $p=q$. Then $G\times H$ is isomorphic to $\Bbb Z_p^2$ which has $p^2-1$ elements of order $p$. So it has $\frac{p^2-1}{p-1}=p+1$ (normal) subgroups of order $p$. $G\times H$ has two other subgroups. $\blacksquare$
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Things look OK, but, if you like, the first half of the proof can be done directly.
If $p\neq q$, $\Bbb Z_p\times \Bbb Z_q\cong \Bbb Z_{pq}$, so it is cyclic. The subgroups of a cyclic group are in correspondence with its divisors, and there are only four divisors of $pq$.
If $p=q$, then your analysis is fine that there are $p^2-1$ elements of order $p$, all of which can be partitioned into the sets that lie in the same subgroup. Each of those sets has $p-1$ elements, so there are only $p+1$ pieces in the partition. That counts all of the nontrivial normal subgroups. With the two trivial subgroups, there would be $p+3$ normal subgroups.