= Solution
The <idele group> is the multiplicative <restricted product>
$$
J_K=\prod_v'K_v^\times
=\{(x_v):x_v\in K_v^\times,\ x_v\in\mathcal O_v^\times\text{ for all but finitely many finite }v\}.
$$
Its <restricted product topology> has basic open sets $\prod_{v\in S}U_v\times\prod_{v\notin S}\mathcal O_v^\times$, where $S$ is finite and contains every infinite place, and $U_v$ is open in $K_v^\times$. In particular, the restriction to almost all local <unit groups> is part of the topology, not merely the unrestricted product topology.
The diagonal map $K^\times\to J_K$ is injective and well-defined by finite support of the <principal ideal>'s <valuations>; its image consists of <principal ideles>. To prove <discreteness of principal ideles>, take the neighbourhood of $1$ which requires $x_v\in\mathcal O_v^\times$ at every finite place and $|x_v-1|<1/2$ in the ordinary real or complex modulus at every infinite place. A diagonal element in it is a global unit, so $x-1\in\mathcal O_K$. If $x\ne1$, the nonzero integer $N_{K/\mathbb Q}(x-1)$ has absolute value at least one, but the archimedean bounds make its absolute value less than $2^{-[K:\mathbb Q]}$. This contradiction proves \b[$K^\times$ is a discrete subgroup of $J_K$].
For a <modulus of a number field>, write $\mathfrak m$ for its finite ideal part together with any selected real places. Let $I_{\mathfrak m}$ be the group of fractional ideals coprime to the finite part. Let $P_{\mathfrak m}$ consist of <principal ideals> $(a)$ with $a\equiv1\pmod{\mathfrak p_v^{m_v}}$ locally at each finite place in the modulus, and $a>0$ at each selected real place. The <ray class group> is
$$
\boxed{\operatorname{Cl}_{\mathfrak m}(K)=I_{\mathfrak m}/P_{\mathfrak m}.}
$$
Equivalently it is the quotient $J_K/(K^\times U_{\mathfrak m})$, with the corresponding principal-unit factors at finite places and positive multiplicative factors at selected real places. The question now assumes that no infinite places occur.
Put $R_{\mathfrak m}=\prod_{v\in S}\mathcal O_v^\times/(1+\pi_v^{m_v}\mathcal O_v)$. Global units map to $R_{\mathfrak m}$ by reduction. To map $R_{\mathfrak m}$ to the <ray class group>, choose, by the <Chinese remainder theorem>, an integral $a$ having the prescribed unit residues at all $v\in S$, and send the tuple to the ray class of $(a)$. A different choice changes the ratio by an element congruent to one at every such place, so the map is well-defined and multiplicative.
The forgetful map to the ordinary <ideal class group> is surjective: <weak approximation for number fields> multiplies an ideal by a <principal ideal> so that all its <valuations> in $S$ vanish. Its kernel consists exactly of classes represented by <principal ideals> coprime to $S$, and their generators supply the local unit residues. Finally, a residue tuple maps to the identity ray class precisely when its chosen $a$ differs from a congruence-one generator by a global unit. Thus the <ray class exact sequence> is
$$
\boxed{\mathcal O_K^\times\longrightarrow R_{\mathfrak m}\longrightarrow\operatorname{Cl}_{\mathfrak m}(K)\longrightarrow\operatorname{Cl}(K)\longrightarrow0.}
$$
The leftmost arrow need not be injective, explaining the absence of an initial zero.
For $K=\mathbb Q(\sqrt2)$, the <ring of integers of Q of square root two> is $\mathbb Z[\sqrt2]$. Its absolute <field norm> is a Euclidean function. Indeed, for $a+b\sqrt2\in K$, choose nearest integers $m,n$; writing $u=a-m$, $w=b-n$, both bounded by $1/2$, gives $|u^2-2w^2|\leq1/2<1$. Applying this to a quotient proves <Euclidean division>, hence the ring is a <principal ideal domain> and $\operatorname{Cl}(K)=0$. This is <norm-Euclideanity of the integers adjoined square root two>.
Write $\mathfrak p=(\sqrt2)$ and $\mathfrak q=(1+2\sqrt2)$. Their <ideal norms> are $2$ and $7$, respectively; $\mathfrak p^2=(2)$ and the <residue field> at $\mathfrak q$ is $\mathbb F_7$, with $\sqrt2\equiv3$ because $1+2\sqrt2\equiv0$. The finite local quotient product for the requested modulus is
$$
R_{\mathfrak m}\cong(\mathbb F_2[t]/(t^2))^\times\times\mathbb F_7^\times\cong C_2\times C_6.
$$
The first factor consists of $1$ and $1+t$.
The given <fundamental unit> $\epsilon=1+\sqrt2$ and $-1$ generate the global units. Their images are
$$
\epsilon\longmapsto(1+t,4),\qquad -1\longmapsto(1,6).
$$
The first image has order six; its cube is $(1+t,1)$ and its square is $(1,2)$, generating the first factor and the order-three subgroup of the second. The second image supplies the missing order-two element of $\mathbb F_7^\times$. Thus the global unit map is surjective onto all $12$ local unit residues. By the exact sequence, or the <unit-surjectivity criterion for a trivial finite ray class group>,
$$
\boxed{\operatorname{Cl}_{2(v)+(v')}(\mathbb Q(\sqrt2))=0.}
$$
There is no positivity condition here: the modulus has no infinite part, so using the unit $-1$ is legitimate and essential in the surjectivity calculation.
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