Proof of $E[y(n)]=E[x(n)] \, \sum h(k)$
The standard meaning of white noise includes an insistence (whether implicit or explicit) that the mean is $0$. Thus, what you want to prove is trivially true: since 
$$Y[n] = \sum_{k=-\infty}^\infty h[n-k]X[k] = h\star X \big\vert_{n},$$ the linearity of expectation (the notion that $E[aX]=aE[X]$ and that the expectation of a sum is the sum of the expectations) gives us that 
$$\mu_Y[n] = E\left[\sum_{k=-\infty}^\infty h[n-k]X[k]\right]
= \sum_{k=-\infty}^\infty h[n-k]E\big[X[k]\big]$$
where all the $E\big[X[k]\big]$ have value $0$, and so both sides have value $0$. 

More generally, if the $X$ process has constant mean $\mu_X$, then 
$$\mu_Y[n] = \sum_{k=-\infty}^\infty h[n-k]E\big[X[k]\big]
= \sum_{k=-\infty}^\infty h[n-k]\mu_X = \mu_X\sum_{k=-\infty}^\infty h[n-k]
= \mu_X \sum_{i=-\infty}^\infty h[i]$$ is also a constant.