Testing whether two groups formed from a continuous variable differ on a binary outcome 18 people, who have been divided into 2 groups based on the semiquantitative expression of a protein (low vs high).
Group 1 (low protein levels) has 6 people, Group 2 (high protein levels) has 12 people. In Group 1, 5 of the 6 people developed a specific disease, whereas only 2 of 12 in Group 2 developed this disease.
I want to determine if there is a statistically significant correlation between low protein levels and the development of this specific disease.
If someone could point me towards the tests I should use, it would be great.
 A: You have a couple of options. One is to do a test (or confidence interval) for the difference of two proportions. Here is a link to a similar question with a solution in R: Change in binomial proportion confidence interval 
Another option is to do a permutation test since your samples are so small. 
A: Taking a step back, why do you want to split the continuous measure of protein levels into a binary variable? Presumably you will lose a lot of information and most likely your predictive accuracy will be reduced.
Unless there are compelling reasons to the contrary, I'd treat protein level as a continuous predictor. You could perform a logistic regression predicting disease outcomes from a continuous measure of protein levels. If the protein coefficient is positive and significant you could make a claim about protein levels being a significant predictor of disease outcome. If you are concerned about the relationship being non-linear you could explore transformations of the protein level variable or explore non-linear predictive relationships (e.g., incorporating a quadratic effect).
A: To perform the permutation test that soakley and Jake recommended in R, use the package exactRankTests. The code in your case is:
library(exactRanktTests)

group1 <- c(rep(1, 5), rep(0, 1))
group2 <- c(rep(1, 2), rep(0, 10))

perm.test(group1, group2, exact=TRUE, conf.int=TRUE)

The $p$-value is $0.0128$ with a 95%-confidence interval ranging from $(0.25, 1.00)$. The $p$-value from prop.test is $0.0262$.
