No Fair Merge via Nondeterminstic Data Flow Streams
When in doubt, consult the original source: Clinger's Thesis. Clinger shows that fair merge cannot be written in a nondeterministic dataflow program. He does this by considering the semantic domains in which such a function would be defined, ultimately deriving a contradiction to the assumption that fair merge exists.

Firstly, as Daniel Apon points out, $merge(\bot,1^\omega)=\{1^\omega\}$. This is the only reasonable definition of merge for these arguments (see page 84 of Clinger's thesis). 

Secondly, $merge(\bot,1^\omega)\subseteq merge(0,1^\omega)$, because  $\bot\le 0$, as stream $0$ is a possible extension of stream $\bot$ (bottom of page 73),  and by assumption $merge$ is monotone. In more detail, the ordering on the domain of strings states that any prefix of a string, is $\le$ the string, thus the empty string $\bot$ is the smallest element, and hence $\bot\le 0$, where $0$ is the string with just the element $0$. The function $merge$ has type $S\times S\to\mathcal{P}(S)$, where $S$ is string and $\mathcal{P}(-)$ is powerset.  The assumption that $merge$ is monotone means that if $a\le b$ then $merge(a,s)\subseteq merge(b,s)$. (Similarly for the second argument.) 

Now $1^\omega\in merge(0,1^\omega)$ violates the assumption that $merge$ is fair – a fair merge of $0$ and $1^\omega$ should to contain the $0$. In more detail, $merge(S,T)$ is interleaving the "cards" from $S$ and $T$, without shuffling them. The significance of the stream $1^\omega$ is that it is infinite. $0$ is simply some character different from $1$. A fair merge of $0$ and $1^\omega$ should result in a string such as $1^*01^\omega$. Fairness means that within some finite time it selects elements from each string, thus the character $0$ is eventually chosen.

Hence $merge$ cannot be written as a nondeterministic dataflow program.