$\newcommand{\cl}{\operatorname{cl}}\newcommand{\int}{\operatorname{int}}$An affirmative answer to (B) follows immediately from the theorem that a locally compact subgroup of a Hausdorff topological group is closed, since $\Bbb R$ is locally compact.
Proof: Let $G$ be a Hausdorff topological group, and let $H$ be a locally compact subgroup of $G$. Let $W$ be a nbhd of $1_G$ in $G$ such that $W\cap H$ is compact, and let $U=\int_GW$. Suppose that $g\in\cl_GH$; then $gU^{-1}$ is an open nbhd of $g$, so we may choose $h\in gU^{-1}\cap H$. Clearly $g\in hU$. Let $$\mathscr{N}=\{N\subseteq G:1\in\int_GN\text{ and }gN\subseteq hU\}\;,$$ and for $N\in\mathscr{N}$ let $H_N=gN\cap H$; $gN$ is a nbhd of $g$, so $H_N\ne\varnothing$ and $g\in\cl_G H_N$. Moreover, $$H_N\subseteq hU\cap H\subseteq hW\cap H=hW\cap hH=h(W\cap H)\;,$$ so $$g\in\bigcap_{N\in\mathscr{N}}\cl_GH_N\subseteq\cl_Gh(W\cap H)\;.$$ $W\cap H$ is compact, so $h(W\cap H)$ is compact and therefore closed in $G$, so in fact we have $g\in h(W\cap H)$. But then in particular $g\in hH=H$, and $H$ is therefore closed. $\dashv$
Added: Martin has provided a negative answer to (A) in the comments below; I take the liberty of repeating it here to make the answer self-contained. Let $\alpha$ be any irrational number, and define $f:\Bbb R\to\Bbb R^2:x\mapsto\alpha x$. Let $q:\Bbb R^2\to\Bbb R^2/\Bbb Z^2$ be the quotient map, and let $h=q\circ f$. Then $h$ is in injective homomorphism of $\Bbb R$ into the torus $\Bbb R^2/\Bbb Z^2$. The fact that $h$ is injective follows from the irrationality of $\alpha$. The irrationality of $\alpha$ also implies that $h[\Bbb R]$ is dense in $\Bbb R^2/\Bbb Z^2$. In fact, pick any $x\in(0,1)$ such that $\alpha x$ is also irrational; this proof that $\{nx:n\in\Bbb Z\}$ is dense in $\Bbb R/\Bbb Z$ can easily be adapted to show that $\{h(nx):n\in\Bbb Z\}$ is dense in $\Bbb R^2/\Bbb Z^2$.
