How to argue about semi-decidability of a problem?
We know that $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$ have the same cardinality, so there exists a bijection $f:\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}$. Let $s_n$ be the $n$-th string in some enumeration, like $s_1=\epsilon, s_2=0, s_3=1, s_4=00, s_5=01,\dotsc$ then consider the following semi-decider, where I leave it to you to fill in the missing parts:
$$\begin{align}
&amp;\Pi'(\Pi_1, \Pi_2)=\\
&amp;\quad\text{for }n=1,2,3,\dotsc\\
&amp;\quad\quad\text{compute }f(n)=(i,j)\\
&amp;\quad\quad\text{run }\Pi_1(s_i)\\
&amp;\quad\quad\text{run }\Pi_2(j)\\
&amp;\quad\quad\text{... }\\
&amp;\quad\quad\text{... }\\
&amp;\quad\quad\quad\text{return accept}
\end{align}$$

There are lots of bijections between $\mathbb{N}$ and $\mathbb{N}\times\mathbb{N}$, here are a few to get you started.