Let $G$ and $H$ be Lie groups and $i:H\to G$ an injective group homomorphism, not necessarily continuous.
Once $i$ is continuous, it follows easily that $i$ is differentiable and even an immersion. So $H$ is a Lie subgroup of $G$ (It is not guaranteed that $H$ is a submanifold of $G$, which is only the case when $i(H)$ is closed in $G$).
However, it is even possible that $i:H\to G$ is not continous?
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It may seem like a silly example, but this kind is the best one can get.
Take $G=\mathbb{R}$ with the discrete topology. It is a Lie group of dimension $0$ with uncountably many connected components.
Take $H=\mathbb{R}$ with the usual structure of Lie group.
The morphism $H\to G$ is not continuous.
It's possible, but the examples are weird.
Consider $H = \mathbb{R}^n$ and $G = \mathbb{R}^m$ with $n\neq m$. If we view both as vector spaces over $\mathbb{Q}$, then we see they are both vector spaces of dimension $|\mathbb{R}|$. It follows that they are isomorphic as $\mathbb{Q}$ vector spaces.
In particular, there is a bijective $\mathbb{Q}$-linear map $i:H\rightarrow G$ with $i(x+y) = i(x) + i(y)$.
Said another way, $i$ is a group homomorphism! It turns out that $i$ is not even measurable, let alone continuous. But, $i(H) = G$, so the image is a Lie subgroup. If you insist on $i(H)$ being a proper subgroup of $G$, embed $G$ into $\mathbb{R}^{m+1}$ in the usual way, then $i:H\rightarrow \mathbb{R}^{m+1}$ has image $\mathbb{R}^m$ embedded in the usual way, but is horribly discontinuous.
On the other hand, if $H$ is compact and semisimple (equivalently, if $H$ is compact and so is the universal cover of $H$) and if $G$ is compact, then van der Waerden proved that any abstract homomorphism $i:H\rightarrow G$ is automatically continuous (hence, smooth).
The reference is