I just wanted to add that there is a fairly easy proof for your final question: Is every continuous homomorphism between Lie groups actually smooth?

The theorem we need is the closed subgroup theorem (also called the Cartan Theorem): If H is a topologically closed subgroup of a Lie group G, then H is actually an embedded Lie subgroup.

Granting this, one proves all continuous homomorphisms are smooth as follows:

Given Lie groups H and G with $f:H\rightarrow G$ a continuous homomorphism, consider the subgroup $K$ of $H\times G$ given by the graph of $f$. The graph is a closed subset of $H\times G$ precisely because $f$ is continuous, and hence, by the closed subgroup theorem, the graph is an embedded smooth submanifold of $H\times G$. Thus, the restriction of the two canonical projection maps $\pi_1:H\times G\rightarrow H$ and $\pi_2:H\times G\rightarrow G$ are smooth when restricted to K.

Now, $\pi_1$ restricted to $K$ is clearly a diffeomorphism onto $H$, and hence has a smooth inverse and so is smooth. But then we find that $f = \pi_2\circ \pi_1^{-1}$ is a composition of smooth maps, and hence is smooth. (To be clearer, the $\pi_1^{-1}$ means the inverse of $\pi_1:K\rightarrow H$.)

I just wanted to add that there is a fairly easy proof for your final question: Is every continuous homomorphism between Lie groups actually smooth?

The theorem we need is the closed subgroup theorem (also called the Cartan Theorem): If H is a topologically closed subgroup of a Lie group G, then H is actually an embedded Lie subgroup.

Granting this, one proves all continuous homomorphisms are smooth as follows:

Given Lie groups H and G with $f:H\rightarrow G$ a continuous homomorphism, consider the subgroup $K$ of $H\times G$ given by the graph of $f$. The graph is a closed subset of $H\times G$ precisely because $f$ is continuous, and hence, by the closed subgroup theorem, the graph is an embedded smooth submanifold of $H\times G$. Thus, the restriction of the two canonical projection maps $\pi_1:H\times G\rightarrow H$ and $\pi_2:H\times G\rightarrow G$ are smooth when restricted to K.

Now, $\pi_1$ restricted to $K$ is clearly* a diffeomorphism onto $H$, and hence has a smooth inverse and so is smooth. But then we find that $f = \pi_2\circ \pi_1^{-1}$ is a composition of smooth maps, and hence is smooth. (To be clearer, the $\pi_1^{-1}$ means the inverse of $\pi_1:K\rightarrow H$.)

*- (Edited in due to comments). One knows by Sard's theorem that there is a point $p\in K$ such that $d_p \pi_1$ is invertible (of full rank). I claim that this implies that for all $q\in K$, $d_q \pi_1$ is invertible. The point is that $\pi_1$ is group homomorphism, which is the same as saying $\pi_1\circ L_{qp^{-1}} = L_{\pi_1(qp^{-1})}\circ \pi_1$, where $L_g$ denotes left multiplication by $g: L_g(h) = gh$. Taking the differentials at p on each side of this equation and using the chain rule, one finds

$$d_q \pi_1 \circ d_p L_{qp^{-1}} = d_{\pi_1(p)}L_{\pi_1(qp^{-1})}\circ d_p \pi_1.$$

The fact that $L_g$ is a diffeomorphism (with inverse $L_{g^{-1}}$) implies that $dL$ is invertible at any point, and hence we see that

$$d_q\pi_1 = d_{\pi_1(p)}L_{\pi_1{qp^{-1}}}\circ d_p \pi_1\circ d_pL_{pq^{-1}};$$ i.e., that $d_q \pi_1$ is a composition of invertible maps, and hence is itself invertible.