For an index set $I$, there shall be topological spaces $(Y_i,\tau_i)$ with a subbase $S_i$ and mappings $f_i:X\rightarrow Y_i$ for any $i\in I$. Let $\tau$ be the initial topology with respect to $(f_i)_{i\in I}$.
Is $S=\bigcup_{i\in I}\{f_i^{-1}(U):U\in S_i\}$ a subbase for the initial topology?
I know that $S'=\bigcup_{i\in I}\{f_i^{-1}(U):U\in \tau_i\}$ is a subbase and clearly $S\subseteq S'$. Furthermore, for any $U_i\in\tau_i$ there are index sets $J_i$ and finite $K_i$ and $U_{j,k}\in S_i$, such that
$$U_i=\bigcup_{j\in J_i}\bigcap_{k\in K_i}U_{j,k}$$
and the preimage could be rewritten as
$$f_i^{-1}(U_i)=\bigcup_{j\in J_i}\bigcap_{k\in K_i}f_i^{-1}(U_{j,k})$$
Let $V$ be a proper open set in the initial topology. Then, I know, there exists an index set $M$, a finite index $N$ and sets $V_{m,n}\in S'$, such that
$$V=\bigcup_{m\in M}\bigcap_{n\in n}V_{m,n}$$
and these can, with the preparation from above (and very sloppy notation), somehow be rewritten as
$$V=\bigcup_{m\in M}\bigcap_{n\in N}\bigcup_{j\in J_i}\bigcap_{k\in K_i}f_i^{-1}(U^{m,n}_{j,k})$$
But now I wonder:
Is this really still a union of finite intersections? Or can it be rewritten as union of finite intersections of elements of S?
Yes. When you write: $$V=\bigcup_{m\in M}\bigcap_{n\in N(m)}\bigcup_{j\in J}\bigcap_{k\in K(j)}f_{i(j,k)}^{-1}(U_{j,k}^{m,n})$$(I've included $f_{i(j,k)}^{-1}$ to show that $i$ will depend, perhaps, on the $j,k$) you've shown $V$ is open in the topology generated by $S$. For each $\bigcap_{k\in K(j)}\cdots$ is open in $S$, so each $\bigcup_{j\in J}\cdots$ is open in $S$, etc.
Another way to realise this would be to endow $X$ with the topology generated by $S$ and ask yourself if this space has the correct characteristic property (functions $Z\to X$ are continuous iff. every $Z\to X\to Y_i$ is continuous). This holds, for a function is continuous if you can show the preimage of any element of the subbase is open.