sigma and mu may or may not be independent depending on the specified priors. In this case you're treating sigma as fixed and only putting a prior on mu, so they are (trivially) independent--a fixed value is independent of every random variable.
x-bar and s are independent for data coming from a Normal distribution, which your model assumes to be the case. Like I said, this is a basic result in mathematical statistics. The first google result [1] for "independence of s and x bar" gives a proof.
> It should also be noted that X and x are different. I defined p(x|mu) to be a Normal distribution but p(X|mu) was never defined to be normal.
From your writeup (apologies for formatting):
"We have N data points X={x_1,x_2,⋯,x_N} independently sampled from normal distribution N(μ,σ^2)."
This means X|mu has a multivariate Normal distribution with mean vector (mu, mu, mu, ..., mu) and covariance matrix sigma^2 * I_n, where I_N is the NxN identity matrix. p(X|mu) is (multivariate) Normal by definition.
Since X|mu has a multivariate Normal distribution, x-bar|mu has a univariate Normal distribution. This is the result of a basic fact about Normal distributions: if a vector X has a multivariate Normal distribution, then MX, where M is any matrix of an appropriate size to be multiplied by X, also has a (multivariate or univariate) Normal distribution (in other words, linear combination of Normally distributed random variables also have a Normal distribution). In the case of x-bar, M is the 1xN matrix [1/N, 1/N, ..., 1/N].
The Central Limit Theorem is a far more general result. In this case since we're starting with a Normal distribution we don't need it, as we can use properties of the Normal distribution to arrive at the distribution of x-bar|mu directly, without needing the CLT.
[1] http://jekyll.math.byuh.edu/courses/m321/handouts/mean_var_i...