Background: The plane cannot be covered by fewer than continuum many lines. The simplest proof I know goes like this:
Suppose $\mathcal{A}$ is a collection of lines in the plane, $|\mathcal{A}|<\mathfrak{c}$. Take a line $l$ not in $\mathcal{A}$. Every line in $\mathcal{A}$ intersects $l$ in at most one point, but there are continuum many points on $\mathcal{l}$. So in fact $\bigcup\mathcal{A}$ misses continuum many points on $l$, and in particular cannot equal $\mathbb{R}^2$.
Note that we don't use any heavy machinery here (or any machinery at all, really) and it also does not depend on how the continuum hypothesis works out.
Question: Does the same hold for injective continuous curves? Specifically, can the plane be covered by fewer than continuum many injective curves, i.e. injective continuous maps $\alpha:\mathbb{R}\rightarrow\mathbb{R}^2$? (Note injectivity is necessary here due to space-filling curves.) The answer is affirmative under CH, but I'm only able to do this using high powered arguments (e.g. noting that the image of each curve has measure zero and the union of countably many null sets is null). But it's consistent for the union of uncountably many but fewer than continuum null sets to have positive measure. (The answer is still affirmative under CH but the argument I just gave doesn't work, see first comment.)
This feels like the sort of question that has been asked before (and in this case either proven true or shown independent, probably in a similar way to cardinal characteristics of the continuum) but I've found it difficult to find references. Thanks!
It's consistent with ZFC that $\operatorname{cov}(\mathcal B)\lt2^{\aleph_0}$, i.e., $\mathbb R$ is the union of fewer than continuum many meager sets.
Suppose $\mathbb R$ is the union of $\kappa$ meager sets. It follows that $\mathbb R$ is the union of $\kappa$ compact nowhere dense sets $A_i$, which are of course totally disconnected, so that $\mathbb R^2$ is the union of $\kappa$ totally disconnected compact sets $A_i\times A_j$. By the Moore–Kline theorem, each of those totally disconnected compact sets is covered by an arc in the plane, i.e., a homeomorph of $[0,1]$. Hence $\mathbb R^2$ is the union of $\kappa$ arcs.