I want to show that let $F \subseteq E$ be an algebraic extension of a field $F$, then the cardinality of $E$ cannot exceed the cardinality of $F[X]$.
The proof this assertion as follows. Let $S$ be the set of all ordered pairs $(f,\alpha)$, where $f\in F[X]$ is nonzero and $\alpha \in E$ with $f(\alpha)=0$. Since for each polynomial f, the number of $\alpha$ such that $(f,\alpha)$ lies in $S$ is finite, we have $|S| \leq \aleph_0|F[X]|=|F[X]|$.
My question is how can I verify the last in equality?
We can write $S$ as a union of the following sets: $$ \bigcup_{f\in F[X]\setminus\{0\}}\{\alpha\in E\colon f(\alpha)=0\}. $$ Each of the sets appearing in the union is finite. Symbolically, we may write $$ \bigl|\{\alpha\in E\colon f(\alpha)=0\}\bigr|<\aleph_0. $$ Now we use the fact that cardinality is subadditive, so that $$ |S|\leq \sum_{f\in F[X]}\aleph_0. $$