Is the Riemann zeta function surjective?

The Riemann zeta function is surjective. First, $\zeta(1/z)$ is holomorphic in the punctured disk $0<|z|<1$. Looking at $z=(1/2+it)^{-1}$ with $t\to\infty$ reveals that $\zeta(1/z)$ has an essential singularity at $z=0$, hence $\zeta(s)$ misses at most one value. If $\zeta(s)=w$ then $\zeta(\overline{s})=\overline{w}$, hence $\zeta(s)$ can only miss a real value. However, $\zeta(s)$ maps the real interval $(1,\infty)$ onto $(1,\infty)$, and the real interval $(-2,1)$ onto $(-\infty,0)$. Also $\zeta(-19)>1$ and $\zeta(-18)=0$, hence $\zeta(s)$ maps $[-19,-18]$ onto a real interval that contains $[0,1]$. So $\zeta(s)$ does not miss any real value, and hence it does not miss any complex value.

$\zeta$ function has only one pole at $z=1$. It also has order $1$. If $\zeta$ omits $c\in C$ then $g:=1/(\zeta-c)$ is entire with one simple zero at $1$. As it is of order $1$, it must be $g(z)=(z-1)e^{az+b}$, by Hadamard's factorization theorem, so $$\zeta(z)=(z-1)^{-1}e^{-az-b}+c,$$ which is absurd.

The formula I wrote is the general form of a non-surjective meromorphic function of order $1$ with a single pole at $1$ and omitting $c$.

The reference on Hadamard's factorization theorem is any texbook on complex variables, for example Ahlfors, Complex Analysis, or Titchmarsh, Function Theory, or Whittaker Watson, or whatever you have.

(Just a quick aside: the set of solutions to the equation $\zeta(s) = a$ when $a \neq 0$ are usually called the $a$-points of the Riemann zeta function in the literature, in case you want to look up the state of the art)

As Steve Huntsman mentions in a comment, it is indeed possible to use the universality of the zeta function to prove that $\zeta(s)$ is surjective with a simple argument based on Rouché's theorem.

In particular, let $U$ be a compact connected subset with a smooth boundary in the critical strip on which universality holds (for simplicity, let $U = \{ z \in \mathbb{C} : |z - 3/4| \leq 1/8 \}$), and let $f:U \to \mathbb{C}$ be a function holomorphic on $U$ (i.e. holomorphic in the interior, and continuous on the boundary) with the following properties:

1. $f$ has no zeroes in $U$.

2. There exists a $z \in U$ such that $f(z) = a$.

3. For all $z \in \partial U$, $f(z) \neq a$.

As an example, for $U$ as above, one could take $f(z) = a + |a| (z-3/4)$.

Now note that 3 implies that there is an $m > 0$ such that $$m = \inf_{z \in \partial U} |f(z) - a|.$$

Pick $\epsilon > 0$ such that $\epsilon < m$. Then, due to 1, universality guarantees that there is a $t \geq 0$ such that $$ |\zeta(s+it) - f(s)| < \epsilon$$ for every $s \in U$. In particular, this inequality holds for every $s \in \partial U$. Thus, for every $s \in \partial U$, we have that

$$ |\zeta(s+it) - f(s)| < \epsilon < m \leq |f(s) - a|. $$

Thus, Rouché's theorem implies that $f(s) - a$ and $f(s) - a + \zeta(s+it) - f(s) = \zeta(s+it) - a$ have the same number of roots in the interior of $U$. Since $f$ has such a root, it follows that $\zeta$ has such a root as well, and we are done.

This argument is robust in a few ways. For one, it applies to other functions which exhibit universality (which is known for fairly widely classes of $L$-functions and other zeta-like families). For another, using more effective versions of Voronin's universality theorem can be used to give a lower bound for the number of $a$-points in a rectangle within the critical strip. For a third, this argument actually proves that there are simple $a$-points for every $a$.

This argument is probably classical, but I first saw it in a paper on simple $a$-points by Gonek, Lester and Milinovich.