The extension $F(X)$ is isomorphic to the field of fractions of the polynomial ring $F[X]$.
What is the cardinality of $F[X]$? Since for every $f\in F[X]$ there exists a finite $X_0\subseteq X$ such that $f\in F[X_0]$, and $\#F[X_0]=\#F$, we have $\begin{align*} \#F[X] &= \#\bigcup_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}F[X_0]\\ &\leq \sum_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}\#F[X_0]\\ &= \sum_{\substack{X_0\subseteq X\\ \#X_0\lt\infty}}\#F\\ &= \#\{X_0\subseteq X\mid \#X_0\lt\infty\}\#F\\ &=\left\{\begin{array}{ll} \#X\#F &\text{if }\ \#X=\infty\\ 2^{\#X}\#F &\text{if }\ \#X\lt\infty \end{array}\right.\\ &= \max\{\#X,\#F\}. \end{align*}$ On the other hand, $F\subseteq F[X]$ and $X\subseteq F[X]$, so $\max\{\#X,\#F\}\leq\#F[X]$. This gives equality.
The field of fractions can be viewed as a quotient of $F[X]\times(F[X]-\{0\})$ modulo the equivalence relation $(f,g)\sim (h,k)\Longleftrightarrow fk=gh$, and since $F[X]$ is infinite, this means that the cardinality is at most that of $F[X]$; since it contains a subring isomorphic to $F[X]$, we conclude that $\#F(X)=\#F[X]$, giving equality.
(A proof that the cardinality of the set of all finite subsets of an infinite set $X$ is equal to that of $X$ can be found here )
On rereading your argument for the first part, it is incomplete, I think.
If your $X$ represents a set of indeterminates, then you need to explain why it cannot have cardinality larger than $F$; if it represents a single indeterminate, then it is false that an algebraic extension must be a surjective image of the polynomial ring in one variable (for example, the algebraic closure of $\mathbb{Q}$ is certainly algebraic over $\mathbb{Q}$, but it is not simple, so it is not the surjective image of $\mathbb{Q}[x]$).
So you need to explain why $\#X\leq\#F$.
A better argument would be to note that there is a map from $K$ to $F[x]$ (single indeterminate) that sends every element of $K$ to its monic irreducible polynomial over $F$. This map is not one-to-one, but each element $f$ of $F[x]$ has at most $\deg(f)$ preimages. As this is finite, it follows that each fiber of the map $K\to F[x]$ is finite. Therefore, $\begin{align*} \#K &\leq \sum_{f\in F[x]}\deg{f}\\ &\leq \sum_{f\in F[x]} \aleph_0\\ & = \#F[x]\aleph_0\\ &=\#F[x]\\ &=\#F \end{align*}$ (since $F$ is infinite). Then noting $F\subseteq K$ gives the other inequality.