Hint: Define
$$f:K[X_1,...,X_n]\to K\;\;,\;\;f(g(X_1,...,X_n)):=g(a_1,...,a_n)$$
1) Show $\,f\,$ is a surjective ring homomorphism
2) Use now the first isomorphism theorem for rings
3) Remember: if $\,R\,$ is a commutative unitary ring, an ideal $\,I\leq R\,$ is maximal iff $\,R/I\,$ is a field.
"Zariski's lemma" states that, for any field $k$, if a field is finitely generated as a $k$-algebra then it is a finite field extension of $k$. Proof of this is below.
However, this means that the circumstances of your problem cannot arise; $A$ is finitely generated as a $k$-algebra (by the $x_i$), and hence – since it is a field – must be a finite and thus algebraic extension of $k$.
Lemma 1: If $E$ and $F=E(x)$ are fields, and $F$ is finitely generated as an $E$-algebra, then $F$ is a finite field extension of $E$.
Proof: Say $F=E[f_1,\ldots,f_n]$, and suppose for contradiction that $F$ is not finite over $E$. Then $x$ must be transcendental over $E$, and so $F$ is the field of rational functions in $x$ over $E$. This means that each $f_i=p_i/q_i$ for some $p_i,q_i\in E[x]$. Consider the element $r:=\frac{1}{x\prod_{i=1}^nq_i+1}\in F$; there must be some polynomial $g\in E[t_1,\ldots,t_n]$ such that $r=g(f_1,\ldots,f_n)$. We can clear denominators in $g(f_1,\ldots,f_n)$ by multiplying through with a suitable power of $\prod_{i=1}^nq_i$, giving $s:=(\prod_{i=1}^nq_i)^kg(f_1,\ldots,f_n)\in E[x]$ for some $k\in\mathbb{N}$, and hence $\frac{1}{x\prod_{i=1}^nq_i+1}=\frac{1}{(\prod_{i=1}^nq_i)^k}s$. But this means $(\prod_{i=1}^nq_i)^k=(x\prod_{i=1}^nq_i+1)s$, and this is a contradiction, since $x\prod_{i=1}^nq_i+1$ and $\prod_{i=1}^nq_i$ (and hence $x\prod_{i=1}^nq_i+1$ and $(\prod_{i=1}^nq_i)^k$) are coprime in $E[x]$.
Lemma 2: Let $A\subseteq B\subseteq C$ rings with $A$ Noetherian. If $C$ is finitely generated as an $A$-algebra and $C$ is finitely generated as a $B$-module, then $B$ is finitely generated as an $A$-algebra.
Proof: See eg here.
Theorem (Zariski's lemma): Suppose $F:E$ is a field extension and that $F$ is finitely generated as an $E$-algebra. Then $F$ is finite over $E$.
Proof: Say $F=E[x_1,\ldots,x_n]$; we prove by induction on $n$. The case $n=0$ is clear. Now suppose the result holds for $n-1$, and let $E'=E(x_n)\subseteq F$. By the tower law, $[F:E]=[F:E'][E':E]$, and so it suffices to show (i) that $F$ is a finite extension of $E'$ and (ii) that $E'$ is a finite extension of $E$. For (i), note that $F=E'[x_1,\ldots,x_{n-1}]$, and so – since $E'$ is a field – by the induction hypothesis we have that $F$ is a finite field extension of $E'$. For (ii), note that (i) allows us to apply lemma 2 and show that $E'$ is finitely generated as an $E$-algebra. By lemma 1, this means that $E'=E(x_n)$ is in fact a finite extension of $E$, so (ii) holds and we are done.
Best Answer
Let me repeat what I said in the comments. There is no ambiguity here. We have the standard notation for the field of fractions $K(x_1,x_2,\dots,x_n)$, and we have a definition of what $K(B)$ means for a set.
The question remains whether $K(x_1,x_2,\dots,x_n)=K(\{x_1,x_2,\dots,x_n\})$. The answer is "of course" and here are some of the details.
Let $K$ be a field, $R:=K[x_1,x_2,\dots, x_n]$ be the polynomial ring in $n$ variables.
Let $L$ denote the field of fractions of $R$, so that $L=\left\{\frac{f}{g}\mid f,g\in R, g\ne0\right\}$. This field is usually denoted by $K(x_1,x_2,\dots,x_n)$.
Now let $B$ be the set $\{x_1,x_2,\dots,x_n\}$. Then consider the field $K(B)$ defined as the smallest subfield of $L$ containing $K$ and the set $B$.
By definition we have that $K(B)\subseteq L$.
Now let $\frac{f}{g}\in L$. We have that $f,g\in K[x_1,x_2,\dots,x_n]$. We can write $f=\sum_{} a_{i_1,i_2,\dots,i_n} x_1^{i_1}x_2^{i_2}\dots x_n^{i_n}$ where the ceofficients are in $K$ and the sum is finite. Now $K(B)$ by definition contains all the $x_i$ and so (being closed under multiplication) contains all the monomials $x_1^{i_1}x_2^{i_2}\dots x_n^{i_n}$. $K(B)$ also by definition contains all the coefficients as these are in $K$. Hence, as $K(B)$ is a field it must contain $f$. The same is true for $g$. But $K(B)$ is a field so it must contain also $\frac{f}{g}$. That is $L\subseteq K(B)$.
We therefore have that $L=K(B)$.
That is, using the usual notation for the field of fractions, $$ K(x_1,x_2,\dots, x_n)=K(\{x_1,x_2,\dots, x_n\}) $$ and we are done.