This was proved by me during the VSRP with a hint from Prof Rajan.

Theorem: Prove that the set of algebraic integers in an algebraic number field forms a ring.

A different Proof:

Denote by $K$, the given algebraic number field and the set of algebraic numbers in $K$, by $\mathcal {O}_K$.

Suppose $\alpha \in \mathcal{O}_K$ satisfies
$f(x)=x^n+a_{n-1} x^{n-1} + \cdots + a_0 \in \mathbb{Z}[x]$.
Then, $( -\alpha )$ satisfies
$f'(x)= (-1)^n a_n x^n + \cdots + (-1)a_1 x + a_0$. Clearly, $f' \in \mathbb Z$ so no problem there.

Now, given $\alpha$ and $\beta$ in $\mathcal{O}_K$, we need to show that $\alpha\beta$ and $\alpha+\beta$ are in $\mathcal{O}_K$.

So assume that $\alpha$ and $\beta$ satisfy the monic polynomials $f(x)$ and $g(x)$ in $\mathbb{Z}[x]$.

Since we can factor $f$ and $g$ in $\mathbb{C}$ using the Fundamental Theorem of Algebra, we can factor $f$ and $g$ as
$f(x) = x^m+s_{m-1} x^{m-1} + \cdots + s_0 = (x-\alpha_1) (x-\alpha_2) \cdots (x-\alpha_m)$

$g(x) = x^n+t_{n-1} x^{n-1} + \cdots + t_0 = (x-\beta_1) (x-\beta_2) \cdots (x-\beta_n)$

where $\alpha=\alpha_1$ and $\beta=\beta_1$ for notational convenience.

Note: The $s_i$‘s and $t_j$‘s are elementary symmetric functions of $\alpha_i$‘s and $\beta_j$‘s respectively.

Now, consider the polynomials
$h(x)= \displaystyle\prod_{i \in \{1,2,\cdots ,m\}} \prod_{j\in \{1,2,\cdots ,n\}} \left(x-\alpha_i \beta_j\right)$

and

$h'(x)=\displaystyle\prod_{i \in \{1,2,\cdots ,m\}} \prod_{j \in \{1,2,\cdots , n\}} (x-(\alpha_i+\beta_j) )$

The polynomials $h$ and $h'$ are invariant under any permutation $\sigma \in S_m$. So, $h$ and $h'$ can be viewed as symmetric polynomials in the variables $\alpha_1, \alpha_2, \cdots, \alpha_m.$ Hence, by the Fundamental Theorem on Symmetric Functions, (not proved!) we have,

$h(x) = h(s_1, s_2, \cdots , s_m) (x)$

Now, since the equation on the right is invariant under any permutation on $n$ characters applied to $\beta_1, \beta_2, \cdots, \beta_n$, it is symmetric in $\beta_j$‘s. So, invoking the theorem on symmetric functions again, we have,

$h(x) = h(s_1, s_2, \cdots , s_m) (x) = h(s_1, s_2, \cdots , s_m) (t_1, t_2, \cdots , t_n) (x)$

Of course, a similar argument holds for $h'$.

Since the coefficients $s_i$‘s and $t_j$‘s are integers, hence $h (x)\in \mathbb{Z}[x]$ and $h' (x) \in \mathbb{Z}[x]$.

Clearly, $\alpha\beta$ and $\alpha+\beta$ are roots of $h$ and $h'$ respectively and hence, $\alpha, \beta \in \mathcal{O}_K. \blacksquare$

Remark: With a little addition, the above proof can be extended to show that the set of all algebraic numbers forms a field.