Assuming the Axiom of Choice, the real line cannot be a union of countably many sets each of size less than continuum. To prove this, one needs to know that the continuum has cofinality strictly greater than $\omega$. (The cofinality of an limit ordinal $\delta$ is the least cardinality of a set $X$ of ordinals strictly less than $\delta$ such that $\delta = \sup (X)$.)
This resolves you question because if $\{ A_n : n \in \omega \}$ is a family of subsets of $\mathbb{R}$ each with size strictly less than $2^{\omega}$, then setting $\kappa = \sup \{ |A_n| : n \in \omega \}$ we have that $\kappa < 2^\omega$. It then follows that $| \textstyle{\bigcup_{n < \omega}} A_n | \leq \sum_{n < \omega} | A_n | \leq \sum_{n < \omega} \kappa = \omega \cdot \kappa = \max \{ \omega , \kappa \} < 2^{\omega},$ and therefore $\bigcup_{n < \omega} A_n$ cannot equal all of $\mathbb{R}$.
Without the Axiom of Choice, it is possible that the continuum is a countable union of countable sets. (Note that this does not contradict the well-known fact (provable without Choice) that $\mathbb{R}$ is uncountable because in such models countable unions of countable sets need not be countable.)