= Solution
For an <adjunction> $L:\mathcal C\rightleftarrows\mathcal D:R$, write its <adjunction unit> as $\eta$ and its <adjunction counit> as $\varepsilon$. It is an <idempotent adjunction> when the induced <monad> has invertible multiplication $\mu=R\varepsilon L:RLRL\Rightarrow RL$. The dual condition is invertibility of the induced <comonad> comultiplication $\delta=L\eta R:LR\Rightarrow LRLR$.
To prove equivalence of these conditions, suppose $\mu$ is invertible. The two unit identities imply $T\eta=\eta T=\mu^{-1}$, where $T=RL$. For any <algebra for a monad> $a:TA\to A$, naturality gives
$$
\eta_Aa=Ta\,\eta_{TA}=Ta\,T\eta_A=T(a\eta_A)=1_{TA},
$$
while $a\eta_A=1_A$. Thus $\eta_A$ is invertible. For each $D\in\mathcal D$, the arrow $R\varepsilon_D:T(RD)\to RD$ is a <monad algebra> action, so $\eta_{RD}$ is invertible. Applying $L$ proves $\delta_D=L\eta_{RD}$ invertible. Apply this implication to the opposite <adjunction>, interchanging the roles of the <monad> and comonad, to obtain the converse. Therefore \b[idempotence is self-dual].
The <open-set frame> $O(X)$ is ordered by inclusion. Its arbitrary <joins> are unions; arbitrary <meets> are interiors of intersections. Finite intersections are already open, so finite <meets> are ordinary intersections, including the empty <meet> $X$. Distributivity of intersection over arbitrary unions proves the <frame> law. If $f:X\to Y$ is a <continuous map>, inverse image
$$
f^{-1}:O(Y)\to O(X)
$$
preserves arbitrary unions and finite intersections, including $\varnothing$ and the entire space. It is therefore a <frame homomorphism>. Since $(gf)^{-1}=f^{-1}g^{-1}$ and $1_X^{-1}=1_{O(X)}$, this defines $O:\mathbf{Top}\to\mathbf{Frm}^{\mathrm{op}}$.
Let $\mathbf2=\{0<1\}$ and let $P(A)$ be the set of <points of a frame> $A$, namely <frame homomorphisms> $A\to\mathbf2$. For $a\in A$, put $U_a=\{p:p(a)=1\}$. Preservation of <joins> and finite <meets> gives
$$
U_0=\varnothing,\qquad U_1=P(A),\qquad U_{a\wedge b}=U_a\cap U_b,\qquad U_{\bigvee_i a_i}=\bigcup_iU_{a_i}.
$$
In the last equality, a <join> in $\mathbf2$ is $1$ precisely when at least one summand is $1$, also covering the empty family. Thus the sets $U_a$ are all the opens of a <topology>, rather than merely a proposed basis.
For a <frame homomorphism> $h:A\to B$, define $P(h):P(B)\to P(A)$ by $p\mapsto p\circ h$. Since $P(h)^{-1}(U_a)=U_{h(a)}$, it is continuous. Composition and identities follow from composition of functions, giving $P:\mathbf{Frm}^{\mathrm{op}}\to\mathbf{Top}$.
The <frame-point adjunction> is the natural <bijection>
$$
\mathbf{Top}(X,P(A))\cong\mathbf{Frm}(A,O(X))
=\mathbf{Frm}^{\mathrm{op}}(O(X),A).
$$
A <continuous map> $f:X\to P(A)$ determines $h_f(a)=f^{-1}(U_a)$; the displayed identities for $U_a$ show that $h_f$ is a <frame homomorphism>. Conversely a <frame homomorphism> $h:A\to O(X)$ determines $f_h(x)(a)=1$ exactly when $x\in h(a)$. Membership in unions and finite intersections proves that $f_h(x)$ is a <frame homomorphism> to $\mathbf2$, and $f_h^{-1}(U_a)=h(a)$ proves continuity. These constructions are inverse. Precomposing in $X$ becomes inverse image of open sets; precomposing a <frame homomorphism> becomes composition of <frame> points. Hence the bijection is natural in both arguments and proves $O\dashv P$.
The <adjunction unit> is
$$
\eta_X:X\to P(O(X)),\qquad \eta_X(x)(V)=1\ \Longleftrightarrow\ x\in V.
$$
The <adjunction counit> at $A$, regarded as an arrow of $\mathbf{Frm}^{\mathrm{op}}$, is represented by the <frame homomorphism> $e_A:A\to O(P(A))$, $a\mapsto U_a$. It is surjective because every open of $P(A)$ is of this form. We can compute $P(e_A):P(O(P(A)))\to P(A)$ explicitly: it sends $q$ to $p=q\circ e_A$. For every open $U_a$ of $P(A)$,
$$
\eta_{P(A)}(p)(U_a)=p(a)=q(U_a).
$$
Thus $\eta_{P(A)}P(e_A)=1$. Conversely, evaluation at $p$ sends $U_a$ to $p(a)$, so $P(e_A)\eta_{P(A)}=1$. Both maps are continuous by their definitions. Therefore $P(e_A)$ is a <homeomorphism>. The multiplication of the induced <monad> $PO$ at $X$ is $P(e_{O(X)})$, which is invertible by this calculation. Consequently \b[the frame-point adjunction is idempotent]. This does not assert that every $\eta_X$ is a <homeomorphism>: it says that applying the point-space construction a second time makes no further change.
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