If $B_b\subset B_a$ and $B_b$ has cardinality less than $c$ for all ordinals $b < a$, does it follow that $\bigcup_{b<a} B_b$ has cardinality less than $c$?
Would it simply "go without saying" that this union has cardinality less than $c$ because, otherwise, there must exist $b$ less than $a$ with $B_b$ not having cardinality less than $c$?
No, this doesn't follow. Fix a strict well-order $\prec$ of $\mathbb R$. Let $( x_{\alpha} \mid \alpha < \mathfrak{c})$ be the strictly $\prec$-increasing enumeration of $\mathbb R$. For $\alpha < \mathfrak{c}$ let $$ B_{\alpha} := \{ x_{\beta} \mid \beta < \alpha \}. $$
Each $B_{\alpha}$ has cardinality $\operatorname{card}(\alpha) \le \alpha < \mathfrak{c}$ and for $\alpha < \beta < \mathfrak{c}$ we have $B_{\alpha} \subseteq B_{\beta}$. But $$ \bigcup_{\alpha < \mathfrak{c}} B_{\alpha} = \mathbb{R} $$ has cardinality $\mathfrak{c}$.