Proof of Existence of Algebraic Closure: Too simple to be true?

You are right. $A$ is a proper class. The reason is simple, by considering all possible fields which are algebraic extensions we immediately have a proper class of sets.

However one can easily observe that if $F$ is a field, then there is a map from $F[x]$ onto any algebraic extension, therefore it suffices to consider algebraic extensions whose underlying set is a partition of $F[x]$.

In either case, one can show that despite the fact that $A$ is a proper class, it is "locally a set", in the sense that below each field there is only set-many fields; and that every chain has size no larger than $|F|+\aleph_0$.

In fact, there is a slightly goosed version (by Jelonek) which avoids the set-theoretic issues. Poles know from set theory...

The other answers rescue your quoted proof of the existence of algebraic closure by observing that there is an explicit bound you can place on the cardinality of the algebraic closure. Here is an alternative proof that side-steps the set-theoretic issue altogether:

Let $S$ be the set of all irreducible polynomials with coefficients in $K$, and let $$R := K\left[\{x_\alpha\}_{\alpha \in S}\right]/\left(\{f(x_\alpha)\}_{\alpha \in S}\right).$$ Then $R$ has a maximal ideal, so we can define $F := R/\mathfrak m$ for $\mathfrak m$ such a maximal ideal, which makes $F$ into an algebraic extension of $K$ in which every polynomial in $S$ has a root.

At this point, it is true that $F$ is algebraically closed, but not easy to prove. You can avoid this additional piece of machinery by simply defining $F_1$ to be the field obtained above, and iteratively defining $F_2, F_3, \ldots$ by the same procedure. A priori, each $f \in S$ can readily be shown to split in $F_i$ for $i \ge \deg f$, and so $F := \displaystyle\bigcup_i F_i$ is an algebraic closure for $K$.