Every field has an algebraic closure proof Below is the proof in Fraleigh.

The part that I don't understand is the construction of set $A$. What does "every possible zero of any $f(x)$" mean? When the textbook talked about "zeros", they are always defined in a larger field, now there is no such single extension fields (at least not proved yet), how should I make sense of such a set?

The set $A$ is not a field it is just a sufficiently large set. For example one could define it slightly more formally as $A = \bigcup \{(f,i)\,|\, f \in F[x];i \in \{0,...,\text{degree}(f)-1\}\}$. Once again for example one could define $\Omega = P(A) \cup F$ (where $P(A)$ denotes the power set of $A$). The idea is that we want a set sufficiently large so that for each algebraic extension we can construct an isomorphic algebraic extension whose underlying set is a subset of this large set. 
