How to prove Cohen forcing does not add unsplit real?

Question

This is an exercise of a forcing lecture notes. The problem is:
Assume $\mathcal{A}$ is an $\omega$-splitting family in $V$. Then $V^\mathbb{C}\models$"$\mathcal{A}$ is $\omega$-splitting".
Here $\mathbb{C}$ is the Cohen forcing. $\mathcal{A}\subseteq[\omega]^\omega$ is $\omega$-splitting means that for every countable sequence $B_i\in[\omega]^\omega$, there is $A\in\mathcal{A}$ such that for all $i$, $|A\bigcap B_i|=|B_i\setminus A|=\omega$.

To prove this, the key lemma is: if $\dot A$ is a $\mathbb{C}$-name for an infinite subset of $\omega$, there exists countable sequence $B_i=B_{\dot A,i}\in [\omega]^\omega$ such that whenever $C\in [\omega]^\omega$ splits all $B_i$, $C$ splits $A$ in the forcing extension.

Let $\mathbb{C}=\{s_i,i\in\omega\}$, I tried to construct $B_i$ as this: $m\in B_i$ iff $\exists p\le s_i(p\Vdash m\in \dot A)$. Does this $B_i$ work? I can't prove it.