Inclusion of multiplicative group of one local field into the idele class group is a closed embedding, but inclusion of more than one isn't?

Let $J^1$ be the ideles of idelic norm one; the global points $k^*$ is contained in $J^1$. In the number field case, we have a surjection $J/k^* \rightarrow \mathbb R _{>0}$ with kernel $J^1/k^*$ (the latter is compact).

Now let $v$ be a place and $O_v^*$ the unit group of $k_v^*$ (if $v$ is complex, $O_v^*=S^1$ ; if $v$ is real, $O_v^*=\{\pm 1\}$; if $v$ is non-archimedean, $O_v^*$ is the group of units). Then under the above quotient map $J/k^*$ the inclusion of $k_v^*$ in $J$ induces a map $k_v^*\rightarrow \mathbb R_{>0}$ whose kernel is the compact group $O_v^* \subset J^1/k^*$. The map $O_v^* \rightarrow J^1/k^*$ is a continuous injection of compact Hausdorff spaces and is hence a homeomorphism onto the image.Since $k_v^*$ is the product of $O_v^*$ with a closed subgroup of $R_{>0}$ (in the archimedean case, it is all of $\mathbb R_{>0}$ and in the non-archimedean case it is a discrete subgroup of $\mathbb R_{>0}$ generated by the uniformizing parameter of $k_v^*$) it follows that we need only check that the map $k_v^* \rightarrow \mathbb R_{>0}$ has closed image. The image is always a closed subgroup and hence if $Card S=1$, $k_v^* \rightarrow J/k^*$ is a closed embedding.

In the non-archimedean case, the image is of the form $\mid \pi \mid ^{\mathbb Z}$ with $\pi $ a uniformizing parameter. If $S$ has two non-archimedean places (which lie above different primes of $\mathbb Q$), the image is of the form $\mid \pi _v \mid ^{\mathbb Z} \mid \pi _w\mid ^{\mathbb Z} \subset \mathbb R_{>0}$ and is not a closed subgroup in $\mathbb R_{>0}$ and hence the embedding of $k_v^*\times k_w^*$ is not closed.

For simplicity, let me assume that $S$ contains the archimedean places. Then the brief answer is that the group of $S$-units is finite when $|S|=1$, but infinite when $|S|>1$. Let me give more detail.

A fundamental system of neighborhoods of the identity in $J$ is provided by the product sets $U=\prod_v U_v$ in $J$, where $U_v$ is an open neighborhood of the identity in $k_v^\times$, and $U_v=r_v^\times$ for all but finitely many $v$'s. Without loss of generality, for any nonarchimedean place, either $U_v=r_v^\times$ or $U_v=1+p_v^{n_v}$ for some $n_v\geq 1$. The corresponding neighborhood in $C$ is $Uk^\times/k^\times$, which intersects the embedded copy of $P$ in $(Pk^\times\cap Uk^\times)/k^\times=(P\cap Uk^\times)k^\times/k^\times$. So in the induced topology of the embedded $P$, a fundamental system of neighborhoods of the identity is provided by the intersections $P\cap Uk^\times$. This intersection consists of those elements $p\in P$, which can be decomposed as $p=ut$ with $u\in U$ and $t\in k^\times$. For any $v\not\in S$, we have that $1=p_v=u_vt_v$, so $t_v=u_v^{-1}\in r_v^\times$. That is, $t$ is an $S$-unit in $k^\times$. Moreover, $t$ is congruent to $1$ modulo a large ideal (determined by the $n_v$'s), but these are all the constraints on $t$. If $|S|=1$, then the group of $S$-units is finite, hence we can force that $t=1$ with an appropriate congruence, which shows that the induced topology on $P$ is the usual one. If $|S|>1$, then the group of $S$-units has rank $|S|-1>0$, which property survives any congruence, so the induced topology will be cruder than the usual one. Indeed, in this case, any $\prod_{v\in S}U_v$ gets translated by infinitely many $t$'s, and only the union of these translates will be open in the induced topology of the embedded $P$, not the translates themselves.