Why do we have to assume normality for a one-sample t-test? As a consequence of the central limit theorem the sampling distribution of the sample means will always be normal whatever is the distribution of the variable we measure.
From our sample we can calculate the standard deviation of the sampling distribution (i.e. standard error) and we can therefore calculate a 95% confidence interval ($\bar x ± 1.96\cdot SE$) in order to accept or reject the null hypothesis.
So, why do we have to assume that the variable is normally distributed? It seems to me that the only thing that has to be normal is the sampling distribution that is approximatively normal given a large enough sample size, isn't it?
 A: The central limit theorem certainly does not guarantee that "the sampling distribution of the sample means will always be normal whatever is the distribution of the variable we measure" in the short run. @Macro has a nice demonstration of how long it can take for the CLT to kick in here: Regression when OLS residuals are not normally distributed.  It may help you to read that (remember that a $t$-test is a simple case of regression).  In short, you need the data to be normal to guarantee that your $p$-values are accurate with your given sample size.  If the data are not normal, your sample size may be adequate, but it may not and it may be difficult for you to know which is true.  
A: Your first sentence might be better written as "As a consequence of the central limit theorem the sampling distribution of the sample means will always approach normality whatever is the distribution of the variable we measure, assuming the variable is distributed finitely and i.i.d.."
A Bernoulli-distributed variable with $p\le0.1$ is pretty much extremely not normal for a finite i.i.d variable (i.e. it takes zero or one as discrete values, rather than any real between $-\infty$ and $\infty$, and is hella skew). Playing with $N$ and decreasing $p$ in the following code may prove instructive about the rate at which the distribution of sampling means approaches normality:
N <- 250
x.bars <- 1:N
for (i in 1:N) {
  x.bars[i] <- rbinom(1,N,0.25)/N
  }
hist(x.bars, breaks=N/10)

In this code, the i.i.d. assumption is given by rbinom(1,N,0.25) so that all "observations" are Bernoulli with $p=0.25$ (i.e. $p$ is constant, and we don't switch distributions depending on the observation), that the possible values are 0s and 1s means we have finite sample distribution, that plus i.i.d. insures we have a finite population distribution.
Here are a series of histograms for $N \in 25, 250, 2500, 25000, 250000, 2500000$ and $p=0.25$:
N=25 (added more bins: N/4)

N=250

N=2500

N=25000

N=250000

N=2500000

There is something really beautiful about that last plot ($N=2.5$m): we can see that there's very much a smooth bell distribution in there... and yet it is also plain that there's something not quite smooth, not perfectly bell about it.
