Let $S$ be a subdirect subgroup of a direct product $G_{1}\times \cdots \times G_{n}$ (that is, each $p_{i}\colon S \mapsto G_{i}$ is surjective) and $p_{i,j}\colon G_{1}\times \cdots \times G_{n} \mapsto G_{i}\times G_{j}$ the projection maps. Suppose that $p_{i,j}(S)$ has finite index in $G_{i}\times G_{j}$ and let $N_{1}=S\cap \text{ker}p_{1}$.
I want to check that $p_{i}(N_{1})$ has finite index in $G_{i}$.
The group $N_{1}$ is the kernel os the restriction to $S$ of $p_{1}=p_{1}\circ p_{1,i}$, but how can I use that $p_{1,i}(S)$ has finite index in $G_{1}\times G_{i}$?
Let $N = \ker p_1$. Since $p_{1,i}(N) = \{(1,g): g \in G_i\}$, we have $$p_{1,i}(N_1) = p_{1,i}(N \cap S) = p_{1,i}(N) \cap p_{1,i}(S),$$ which has finite index in $p_{1,i}(N) \cap (G_1 \times G_i) = \{(1,g):g \in G_i\}$. So $p_i(N_1) = p_i'(p_{1,i}(N_1))$ has finite index in $p_i'(\{(1,g):g \in G_i\}) = G_i$, where $p_i'$ is the projection from $G_1 \times G_i$ to $G_i$.