Real Dirichlet characters mod $k$ How to show that for a non-principal real Dirichlet character mod $k$, there are infinitely many primes $q$ such that $\chi(q)=1$ and infinitely many primes $q'$ such that $\chi(q')=-1$.

As others have pointed out, this is easy to prove with Dirichlet's theorem. However, since you used the (analytic-number-theory) tag, here's a proof from first principles:
According to a the "main lemma" of Dirichlet, for any nontrivial character $\chi$, the $L$-function $L(s, \chi)$ is nonzero and holomorphic at $s=1$.
Hence this means $$0 <\prod_{p}(1-\chi(p)p^{-1})^{-1} < \infty$$
hence 
$$-\infty < \sum_{p}\log(1-\chi(p)p^{-1}) = \sum_{p}\sum_{n\geq 1}\frac{\chi(p^n)}{np^n}< \infty.$$
Now the only non-negligible thing in this sum is the sum of the terms when $n=1$, i.e.
$$\sum_{p}\frac{\chi(p)}{p}.$$
Thus, 
$$\left|\sum_{p}\frac{\chi(p)}{p}\right|<\infty.$$
Now since $$\sum_{p}\frac{1}{p}$$ diverges, there must be infinitely many $p$ for which $\chi(p)=1$ and infinitely many $p$ for which $\chi(p)=-1$. (Note that there are only finitely many $p$ for which $\chi(p)=0$.) $\: \: \square$
This observation leads to the formulation of one of the great theorems of modern number theory, the Chebotarev density theorem.
Addendum: here is a plot of the partial sums $\displaystyle \sum_{p \leq p_n}\frac{\chi(p)}{p}$ for $\chi = \left(\frac{\cdot}{5}\right)$:

The behaviour is quite complex and gives an idea of the rather uniform distribution of $\chi(p)$. For comparison, here is an example of partial sums $\displaystyle \sum_{p \leq p_n}\frac{f(p)}{p}$ where $f$ takes random uniformly distributed values in $\{-1,1\}$:

And finally, here is a picture of a baby duck.
