Is a random sample of a Poisson distribution also Poisson distributed? Car analogy:
Assume the traffic (number of cars per hour) on a road has a Poisson distribution, and the time between cars has the matching exponential distribution. If the chance of each car being red is independent from both time and the color of other cars, will the number of red cars per hour also have a Poisson distribution? 
I strongly suspect so. Furthermore I expect E(red cars)=p(red) * E(cars), and because of the Poisson distribution σ(red cars) = p(red) * σ(cars). But how would I (dis)prove this?
 A: The answer to your first question is yes. If the sum of two independent variables is poisson then the individual variables are also poisson. See Raikov's Theorem 
A: What you have described is a standard and important result from the theory of stochastic processes. Pick up any good book in the library on those, flip to the "Poisson Processes" chapter, and look for "decomposition" or sometimes "thinning".
Two of my favorites are Introduction to Stochastic Processes by Cinlar and Essentials of Stochastic Processes by Durrett.
A: I think you need additional assumptions to draw this conclusion.
Imagine this: the cars on the road are originally unpainted, but just before you observe them a demon decides what color they will become: red or some other color (of which there could be many).  The demon does not have a clock (to ensure that whatever he does is independent of time), but he does have a fair coin.  Ordinarily he flips the coin to determine whether the car will be red or some other color.  However, his can of red paint can color ten cars and needs to be used as quickly as possible.  So, if the coin decides red, the demon just goes ahead and paints the next ten cars red.  Once the paint is exhausted, he goes back to flipping the coin.  Other colors are painted in the same fashion.
Obviously the car count per unit time is unaffected by this procedure, but I hope it's just as obvious that the count of red cars per unit time is not a Poisson process, because it will be overdispersed: there will be too many high counts, due to the temporal clumping of red cars, and too many low counts, due to the temporal clumping of non-red cars.
It could be that this scenario violates your sense of "the chance of red being independent of the color of the other cars," but it's difficult to know exactly what this statement means, since it's open to various interpretations.  In the present case, the probability that the next car to be observed is red, given that the previous car was non-red ("the other" colors), is independent of the color of the previous car.
A: you could model the traffic flow of red cars as a compound poisson distribution.
suppose we look at the number N of cars that pass in a fixed time [8 hours, perhaps].
N should have a poisson distribution, according to your assumptions. let $\lambda$ be the poisson parameter for N. 
suppose all of the cars are colorless to begin with. as each car passes, we flip a coin for which P(heads) = $p$. [presumably $p$ will reflect the proportion of red cars in the population.] each time the coin lands 'heads' the car is colored red.
then if M denotes the number of red cars seen in the time period involved, M is also poisson with parameter $p\lambda$. [easy homework exercise, or see feller's book: intro to probability theory$\cdots$, vol 1, wiley.]
