$\let\sset\subseteq\def\cF{\mathcal F}\def\R{\mathbb R}\def\Q{\mathbb Q}$The answer to both questions is negative: let $\preceq$ be a well order of type $\omega_1$ on a subset of $\R$, and let $\cF$ consist of all proper initial segments of $\preceq$. Then $\cF$ is a family of countable sets totally ordered by $\subseteq$, but $\bigcup\cF$ is uncountable, and $(\cF,{\sset})$ does not embed in $(\R,{\le})$ as $(\omega_1,{\le})$ does not embed there.

In fact, for any $\cF$ as in the question, the following are equivalent:

$(\cF,{\sset})$ embeds in $(\R,{\le})$.

$\bigcup\cF$ is countable.

$(\cF,{\sset})$ has countable cofinality.

1 → 3: Any subset of $\R$ has a countable cofinal subset.

3 → 2: If $C\sset\cF$ is a cofinal countable subset, then $\bigcup\cF=\bigcup C$ is a countable union of countable sets, hence countable.

2 → 1: Let $I=\bigcup\cF$. Then $\{(x,y):\forall U\in\cF\,(y\in U\to x\in U)\}$ is a total preorder on $I$, hence it includes a total order $\preceq$ on $I$, and every $U\in\cF$ is an initial segment of $\preceq$. Since the lexicographic product $(I,{\preceq})\times2$ embeds in $(\Q,{\le})$, we can find an embedding $f\colon(I,{\preceq})\to(\Q,{\le})$ such that $f(x)>\sup\{f(y):y\prec x\}$ for all $x\in I$. Then $U\mapsto\sup f[U]$ is an embedding of $(\cF,{\sset})$ in $(\R,{\le})$.

Note that the example in the beginning is, in a sense, the worst that can happen: as any total order, $(\cF,{\sset})$ has a well-ordered cofinal subset $C$. If $C$ is not countable, it can only have order type $\omega_1$, as otherwise some element of $\cF$ is uncountable. Thus, $(\cF,{\sset})$ has cofinality $\omega_1$, $|\bigcup\cF|=\aleph_1$, and using a similar argument as above, $(\cF,{\sset})$ embeds in the long line $\omega_1\times[0,1)$.

7more comments