Here is a spherical-cow physicist's estimate. Beware: spherical-cow physicist's estimates are extremely dangerous when dealing with climate: for instance the obvious model of how the greenhouse works that physicists (including me) have is simply wrong: it's not approximately right, it's wrong, and it totally mispredicts the effect. So be very wary about numbers below: they could be out by orders of magnitude.
OK, so first of all let's model the Earth as a black body and compute the change in flux for a given temperature change (this is probably the most egregious bit of spherical-cowism here).
So flux is given by
$$F = \sigma T^4$$
And so
$$
\begin{align}
F + \Delta F &= \sigma (T + \Delta T)^4\\
&= \sigma T^4\left(1 + \frac{\Delta T}{T}\right)^4\\
&= \left(1 + \frac{\Delta T}{T}\right)^4F\\
&\approx \left(1 + \frac{4\Delta T}{T}\right)F
\end{align}
$$
Where in the last line I've assumed $\Delta T \ll T$.
And this gives us
$$\frac{\Delta F}{F} \approx \frac{4\Delta T}{T}$$
But $\Delta F/F = \Delta A/A$ where A is the area of the Earth that is intercepting the Sun's light (this is another spherical-cowism to some extent), and hence
$$
\begin{align}
\Delta A &= A\frac{\Delta F}{F}\\
&\approx 4A\frac{\Delta T}{T}\\
&\approx 4\pi R^2 \frac{\Delta T}{T}
\end{align}
$$
Where $R$ is the radius of the Earth.
So, plug in the numbers: $T\approx 287\,\mathrm{K}$, $\Delta T = -1\,\mathrm{K}$, $R \approx 6.37\times 10^6\,\mathrm{m}$, and we get
$$\Delta A \approx -1.78\times 10^{12}\,\mathrm{m^2}$$
If this is to be in orbit, then even with the rotating mirror trick we need to double it, so we need to fly an area of about $3.55\times 10^{12}\,\mathrm{m^2}$.
OK, what will the mass of this be? Let's make it from aluminium and let's make it $0.1\,\mathrm{mm}$ thick (so, not foil, but this accounts for all the structure to make it rotate as well). The mass of this is
$$3.55\times 10^{12}\,\mathrm{m^2}\times 10^{-4}\,\mathrm{m}\times 2700\,\mathrm{kg/m^3} \approx 4.8 \times 10^{11}\,\mathrm{kg}$$
Let's assume they can lift $2\times 10^5\,\mathrm{kg}$ to LEO (200 tonnes), which is more than they can by a fair factor. So this is
$$\frac{4.8 \times 10^{11}}{2 \times 10^5} \approx 2.4\times 10^6$$
launches.
Let's say I'm out by a factor of 100 – I have the mirrors too heavy by a factor of 10 and too large by a factor of 10. So that brings it down to a mere 24,000 launches. If they can launch one of these things every day then 24,000 launches takes 65 years. And I'll refer you to an older answer of mine for some calculations about the climate cost of launches: 24,000 might not be too bad, but millions would be awful, and 24,000 is very, very low estimate I think.
So, not even slightly plausible. Almost certainly not even slightly plausible to mine the stuff on the Moon and lift it from there.
A note on geoengineering
Nothing in this answer or in my replies to comments below should be taken as meaning I think that this or any other form of geoengineering is a good solution to climate change. Any form of geoengineering is clearly a last-ditch thing which is fraught with terrifying risks and uncertainties. However if we do want to (or have to) to do geoengineering then lifting half a billion tonnes of material into LEO seems to me a particularly mad approach due to the enormous energy (and therefore carbon!) cost of getting material into orbit.