Proof for symmetric polynomials with two variables X, Y:
Let $f(X,Y)$ be a symmetric polynomial with degree n. By definition, $f(X,Y)=f(Y,X)$.
Now, $$f(X,Y)= \sum_{0 \le i\le n,0\le j \le n,i+j\le n} a_{ij}X^iY^j$$
for some $a_{ij}$, and $a_{ij}=a_{ji} \;\forall{0\le i \le n,0\le j\le n}.\;\;\;\;(*)$
If $\alpha,\beta \in{\mathbb R}, c$ are some arbitrary constants, and $f(X,Y), g(X,Y)$ are two symmetric polynomials, then $\alpha f(X,Y)+\beta g(X,Y)+c$ is also a symmetric polynomial. This obviously extends to cases with multiple symmetric functions, i.e.
$$\sum_{j=1}^{p} \alpha_j f_j(X,Y)+c \; \text{is symmetric} \;\forall{\alpha_j \in{\mathbb R}, f_j \;\text{are all symmetric}} \;\;\;\;(**)$$
If n is even, then let $n=2m$
$$(X+Y)^n=X^n+\binom n 1 X^{n-1}Y+ \;...\;+\binom n mX^mY^m+ \;...\;+\binom n {n-1} XY^{n-1}+Y^n$$
If n is odd, then let $n=2m+1$
$$(X+Y)^n=X^n+\binom n 1 X^{n-1}Y+ \;...\;+\binom n mX^{m+1}Y^m+\binom n {m+1}X^mY^{m+1}+ \;...\;+\binom n {n-1} XY^{n-1}+Y^n$$
It is clear that $(X+Y)^n$ is symmetric $\forall{n\in {\mathbb N}}$. Moreover, $\forall 0\le k\le n,$ the $k^{th}$ term is the same as the $(n+2-k)^{th}$ term.
By $(*)$ and $(**)$ , it suffices to prove that every symmetric polynomial $f_0(X,Y)$ with at most 2 non-constant terms can be generated by $X+Y$ and $XY$, and that all polynomials with at most 2 non-constant terms generated by $X+Y$ and $XY$ are symmetric.
Suppose $\mu X^iY^j$ is symmetric, where $i\ge 0, j\ge 0$ are not both $0$. Then $X^iY^j=X^jY^i,$ which implies $i=j$.
Now, let us consider the polynomial $\;\eta_1 X^iY^j + \eta_2 X^rY^s$, where $i+j \ne r+s, i+j \ne 0, r+s \ne 0$. If it is symmetric, then $$\;\eta_1 X^iY^j + \eta_2 X^rY^s=\;\eta_1 X^jY^i + \eta_2 X^sY^r$$
This means either $X^iY^j=X^jY^i,X^rY^s=X^sY^r$ or $\eta_1=\eta_2, X^iY^j=X^sY^r, X^jY^i=X^rY^s$. Therefore, in the first case, $i=j, r=s$, and in the second case, $i=s,j=r$. Hence, all symmetric polynomials with two non-constant terms are of the form $\eta X^iY^i+\omega X^jY^j+c \;\text{or}\; \gamma (X^iY^j+X^jY^i)+c$.
Now, $\mu X^iY^i$ and $\eta X^iY^i+\omega X^jY^j$ can clearly be generated by $XY$, so we only have to check $\gamma (X^iY^j+X^jY^i)\;,$ where $i\ne j$.
Without loss of generality, suppose that $i\gt j$, so we have $$\gamma (X^iY^j+X^jY^i)=\gamma X^jY^j(X^{i-j}+Y^{i-j})$$. Now we only have to check if $$X^{i-j}+Y^{i-j}$$ can be generated by $X+Y$ and $XY$.
Let $i-j=q$. When $q=1$, $X^q+Y^q$ can trivially be generated by $X+Y$, since it's itself $X+Y$. When $q=2, X^q+Y^q=X^2+Y^2=(X+Y)^2-2XY,$ so $X^q+Y^q$ can be generated by $X+Y$ and $XY$ as well.
Notice that
$$X^{q+1}+Y^{q+1}=(X^q+Y^q)(X+Y)-(XY^q+YX^q)=(X^q+Y^q)(X+Y)-XY(X^{q-1}+Y^{q-1})$$
Therefore, $X^{q+1}+Y^{q+1}$ can be generated by $X+Y,XY,X^{q-1}+Y^{q-1},$ and $X^q+Y^q$. Hence, by induction,$\forall{q\in {\mathbb N}}, X^q+Y^q$ can be generated by $X+Y$ and $XY$. This means
$$\gamma(X^iY^j+X^jY^i)$$ can be generated by $X+Y$ and $XY$, hence all symmetric polynomials $f(X,Y)$ can be generated by $X+Y$ and $XY$.
All polynomials of degree n generated by $X+Y$ and $XY$ are of the form
$$\sum_{0 \le i'\le n,0\le j' \le n,i'+j'\le n} \kappa_{i'j'}(X+Y)^{i'}(XY)^{j'}$$
Each term of this sum is symmetric, so the sum is symmetric as well. Hence, every polynomial generated by $X+Y$ and $XY$ is symmetric, and every symmetric polynomial can be generated by $X+Y$ and $XY$.
Let $\tau_1, \tau_2,\dots,\tau_{n-1}$ be the elementary symmetric polynomials in $x_2,x_3,\dots,x_n$. It is straightforward to show that $\sigma_k=\tau_k + x_1 \cdot \tau_{k-1}$ for $k=1,2,\dots,n-1$ (with $\tau_0=1\,$).
It follows that each $\tau_k$ can be written as a polynomial $t_k$ in $x_1$ and $\sigma_1, \sigma_2,\dots,\sigma_{n-1}$:
$$
\begin{align}
\tau_1 &= \sigma_1 - x_1\,\tau_0 = \sigma_1 - x_1 = t_1(x_1, \sigma_1)
\\ \tau_2 &= \sigma_2 - x_1\,\tau_1= \sigma_2 - x_1\,t_1(x_1,\sigma_1) = t_2(x_1, \sigma_1, \sigma_2)
\\ \dots
\\ \tau_{n-1} &= \sigma_{n-1} - x_1 \, \tau_{n-2} = \sigma_{n-1} - x_1 \, t_{n-2}(x_1, \sigma_1,\dots,\sigma_{n-2}) = t_{n-1}(x_1, \sigma_1,\dots,\sigma_{n-1})
\end{align}
$$
A polynomial that is symmetric under permutations of the $x_i$’s that fix $x_1$ can then be written as:
$$
\begin{align}
f(x) &= \sum_{j=0}^n \;p_j(\tau_1, \tau_2, \dots,\tau_{n-1})\,x_1^j
\\ &= \sum_{j=0}^n \;p_j\left(t_1(x_1, \sigma_1), t_2(x_1, \sigma_1,\sigma_2), \dots,t_{n-1}(x_1, \sigma_1,\sigma_2,\dots,\sigma_{n-1})\right)\,x_1^j
\\ &= p(x_1, \sigma_1,\sigma_2,\dots,\sigma_{n-1})
\end{align}
$$
This proves the proposition for polynomials, and the extension to rational functions follows.
Best Answer
Let $f\in K[X_1,\dots,X_n]$ be a symmetric polynomial. Then $$f\in K(X_1,\dots,X_n)^{S_n}=K(\sigma_1,\dots,\sigma_n).$$ We want to prove that $f\in K[\sigma_1,\dots,\sigma_n]$. The ring extension $K[\sigma_1,\dots,\sigma_n]\subset K[X_1,\dots,X_n]$ is integral since $X_i$ is integral over $K[\sigma_1,\dots,\sigma_n]$ for all $i=1,\dots,n$. (Note that $X_i$ is a root of the monic polynomial $X^n − \sigma_1X^{n−1} +\cdots +(−1)^n\sigma_n\in K[\sigma_1,\dots,\sigma_n]$.) In particular, $f$ is integral over $K[\sigma_1,\dots,\sigma_n]$. Since $f\in K(\sigma_1,\dots,\sigma_n)$ and $K[\sigma_1,\dots,\sigma_n]$ is integrally closed (why?) we get $f\in K[\sigma_1,\dots,\sigma_n]$.