Some details in the proof of existence of algebraic closure

Solution 1:

Basically what you are asking is why $L:=K[\textbf{X}]/M$ (where $\textbf{X},M$ are defined in item 1 and 3 in your question) is an extension of $K$ generated by a set of algebraic elements over $K$.

(by taking $(L_n,L_{n+1})=(K,L)$).

Here is an outline :

1. It suffices to show that the image $x_f$ of each $X_f\in\textbf{X}$ in $L$ is algebraic over $K$.
2. By the definition of $M$, each $x_f$ is evidently algebraic over $K$.

To elaborate a little more on this :

1. is to use the fact that $L$ is generated by $x_f$ over $K$ ($L$ being a quotient of $K[\textbf{X}]$).

2. is by the definition of $M$ : for each $f$, one has $f(x_f)=0$ in $L$.

Notice that we don't need $M$ (or the ideal $I$ constructed in your question) to be finitely generated in $K[\textbf{X}]$ in these lines.