Lie subgroups are certainly not always embedded (there is the example of the $\mathbb{R} \to S^1 \times S^1$ given by a line of irrational slope).
Can you have a torus that is a subgroup of a Lie group, but not embedded?
To me it seems like the image of a compact set is compact and hence closed (since manifolds are Hausdorff) if the inclusion is continuous. So we want a torus subgroup included in a Lie group in a non-continuous way.
I really have no idea how to come up with such an example.
Let's consider the torus $T=S^1$. As an abstract group, it is isomorphic to $$ {\mathbb Q}/{\mathbb Z} \times \bigoplus_{t\in {\mathbb R}} {\mathbb Q}. $$ Therefore, it contains a proper subgroup
$$ T'= {\mathbb Q}/{\mathbb Z} \times \bigoplus_{t\in {\mathbb R} -\{0\} } {\mathbb Q}. $$ Clearly, $T'$ is isomorphic to $T$ as an abstract group. However, if we equip $T$ with the standard topology of $S^1$ (making it a Lie group) then $T'$ cannot be a Lie subgroup of $T$. Thus, you get an example of a subgroup of a Lie group which is isomorphic to a torus (as an abstract group) but is not a Lie subgroup.