$F$ is a subfield of $\mathbb{R}$, if $F$ is uncountable, then $F=\mathbb{R}$?
$F$ is a proper subfield of $\mathbb{R}$(as a topological field), if $F$ is uncountable, then $F$ is not locally compact?
3.Does there exist a connected, non-locally compact, Hausdorff topological division ring (or field)?
(1) No; for instance, if $S\subset\mathbb{R}$ is an uncountable algebraically independent set, then $F=\mathbb{Q}(S)$ does not contain any irrational algebraic numbers but is uncountable.
(2) Yes; if a subfield $F\subseteq\mathbb{R}$ is locally compact, then for some $\epsilon>0$, $F\cap[-\epsilon,\epsilon]$ is compact and hence closed in $\mathbb{R}$. But $F$ contains $\mathbb{Q}$, so this means $[-\epsilon,\epsilon]\subset F$. Since $F$ is closed under addition, this implies $F=\mathbb{R}$.
(3) Yes. For instance, several topologies on $\mathbb{C}(t)$ compatible with the field structure are discussed in this paper of J. H. Williamson. These topologies also make $\mathbb{C}(t)$ a topological vector space, so it is connected (and, being an infinite-dimensional topological vector space, it is not locally compact). The simplest of these topologies is the following: consider each element of $\mathbb{C}(t)$ as a function $(0,1)\to\mathbb{C}\cup\{\infty\}$, and give it the topology of convergence in measure. Explicitly, this means that a basis is given by sets of the form $$\{g\in\mathbb{C}(t):\mu(\{x\in(0,1):|g(x)-f(x)|>N\})<\epsilon\}$$ where $f\in\mathbb{C}(t)$, $N,\epsilon>0$, and $\mu$ denotes Lebesgue measure. It is not hard to check that addition, multiplication, and inversion are all continuous with respect to this topology, and the details are written out on page 731 of the paper.