If $K$ is a number field, and there exists $\alpha$ such that $\mathcal{O}_K=\mathbb{Z}[\alpha]$, then $K$ is called monogenic. The first example of a non-monogenic number field is due to Dedekind, who showed that for $K=\mathbb{Q}(\theta)$ where $\theta$ is a root of the cubic polynomial $x^3-x^2-2x-8$, the ring of integers does not satisfy $\mathcal{O}_K=\mathbb{Z}[\delta]$ for any $\delta$. In what follows, I will provide a full proof of this claim.
Proof that $\mathcal{O}_K\neq \mathbb{Z}[\delta]$ for any $\delta$: First, we verify that $f$ is indeed irreducible over $\mathbb{Q}$.
Since it is a cubic polynomial, if it were reducible it would have
a rational root which is impossible by the rational root theorem.
Let $\eta=\frac{\theta^{2}+\theta}{2}$, then by calculating the determinant
and traces, note that $\eta^{3}-3\eta^{2}-10\eta-8=0$. The elements$1,\theta,\eta$
are independent over $\mathbb{Q}$ since $\theta$ does not satisfy
a degree $2$ equation, and so $\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}$ is a full rank subring of $\mathcal{O}_{K}$.
I claim that $\mathcal{O}_{K}=\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}$.
To prove this, we calculate the discriminant.
Let
$$
B=\left[\begin{array}{ccc}
\text{Tr}_{K/\mathbb{Q}}\left(1\right) & \text{Tr}_{K/\mathbb{Q}}\left(\theta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\eta\right)\\
\text{Tr}_{K/\mathbb{Q}}\left(\theta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\theta^{2}\right) & \text{Tr}_{K/\mathbb{Q}}\left(\theta\eta\right)\\
\text{Tr}_{K/\mathbb{Q}}\left(\eta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\eta\theta\right) & \text{Tr}_{K/\mathbb{Q}}\left(\eta^{2}\right)
\end{array}\right].
$$
Writing $\theta$ and $\eta$ in the basis $1,\theta,\theta^{2}$,
we have that
$$
M_{\eta}=\left[\begin{array}{ccc}
0 & \frac{1}{2} & \frac{1}{2}\\
4 & 1 & 1\\
8 & 6 & 2
\end{array}\right],\ M_{\theta}=\left[\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
8 & 2 & 1
\end{array}\right],
$$
and from this we find that
$$
M_{\theta}=\left[\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
8 & 2 & 1
\end{array}\right],\ M_{\theta}^{2}=\left[\begin{array}{ccc}
0 & 0 & 1\\
8 & 2 & 1\\
8 & 10 & 3
\end{array}\right],\ M_{\theta}M_{\eta}=\left[\begin{array}{ccc}
4 & 1 & 1\\
8 & 6 & 2\\
16 & 12 & 8
\end{array}\right],\ M_{\eta}^{2}=\left[\begin{array}{ccc}
6 & \frac{7}{2} & \frac{3}{2}\\
12 & 9 & 5\\
40 & 22 & 14
\end{array}\right].
$$
It follows that $\text{Tr}_{K/\mathbb{Q}}\left(1\right)=3$, $\text{Tr}_{K/\mathbb{Q}}\left(\theta\right)=1$,
$\text{Tr}_{K/\mathbb{Q}}\left(\theta^{2}\right)=5$, $\text{Tr}_{K/\mathbb{Q}}\left(\eta\right)=3$,
$\text{Tr}_{K/\mathbb{Q}}\left(\eta^{2}\right)=29$, $\text{Tr}_{K/\mathbb{Q}}\left(\theta\eta\right)=18$,
and so
$$
B=\left[\begin{array}{ccc}
3 & 1 & 3\\
1 & 5 & 18\\
3 & 18 & 29
\end{array}\right].
$$
Thus
$$
\text{disc}_{K/\mathbb{Q}}\left(\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}\right)=\det B=-503.
$$
As $503$ is a prime number, this implies that $\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}$
equals the ring of integers, since otherwise we must have $\text{disc}_{K/\mathbb{Q}}\left(\mathbb{Z}\oplus\theta\mathbb{Z}\oplus\eta\mathbb{Z}\right)=m^{2}\text{disc}_{K/\mathbb{Q}}\left(\mathcal{O}_{K}\right)$
for some integer $m>1$.
Now, let $\delta=a+b\theta+c\eta$ be a general element of $\mathcal{O}_{K}$.
Our goal is to show that $2|\text{disc}\left(\mathbb{Z}[\delta]\right)$.
The matrix for $M_{\delta}$ in the basis $1,\theta,\eta$, equals
$$
M_{\delta}=\left[\begin{array}{ccc}
a & b & c\\
4c & a-b & 2b+2c\\
4b+6c & 2c & a+2b+3c
\end{array}\right].
$$
Reducing modulo $2$, we have that
$$
M_{\delta}\equiv\left[\begin{array}{ccc}
a & b & c\\
0 & a-b & 0\\
0 & 0 & a+c
\end{array}\right]\text{ mod }2.
$$
Since this upper triangular, it follows that
$$
\text{Tr}\left(M_{\delta}^{k}\right)\equiv a^{k}+(a-b)^{k}+(a+c)^{k}\text{ mod }2.
$$
$$
\equiv a-b+c\text{ mod 2}.
$$
Now, if $a-b+c\equiv0\text{ mod }2$, then the last column in the
matrix
$$
A=\left[\begin{array}{ccc}
\text{Tr}\left(1\right) & \text{Tr}\left(\delta\right) & \text{Tr}\left(\delta^{2}\right)\\
\text{Tr}\left(\delta\right) & \text{Tr}\left(\delta^{2}\right) & \text{Tr}\left(\delta^{3}\right)\\
\text{Tr}\left(\delta^{2}\right) & \text{Tr}\left(\delta^{3}\right) & \text{Tr}\left(\delta^{4}\right)
\end{array}\right]
$$
has each entry divisible by $2$, and hence $\text{disc}_{K/\mathbb{Q}}\left(\mathbb{Z}[\delta]\right)=\det A$
is divisible by $2$. If $a-b+c\equiv1\text{ mod }2$, then every
element in $A$ is an odd number. As this determinant is a sum over
permutations in $S_{3}$, we see that we are summing exactly $3!=6$
odd numbers, and so the determinant is even. In either case it follows
that $2|\det A$ as well, and since $\text{disc}_{K/\mathbb{Q}}\left(\mathcal{O}_{K}\right)=-503$,
we have shown that $\mathcal{O}_{K}\neq\mathbb{Z}[\delta]$ for any
$\delta\in\mathcal{O}_{K}$ .
Best Answer
Note that in this case
$$\mathcal O_L\subseteq d^{-1}\alpha_1\mathcal O_K+\cdots+d^{-1}\alpha_n\mathcal O_K$$
and the latter is a free $\mathcal O_K$-module and in particular finitely generated. We may consider the special case $K=\Bbb Q$ and $\mathcal O_K=\Bbb Z$. Then, $\mathcal O_L$ is as a submodule of a finitely generated module over a Noetherian ring is itself finitely generated. This by passes the usual trace argument in so far as this argument is hidden in the lemma. Using this on $\mathcal O_K\subseteq\mathcal O_L$ for a finite extension of number fields $L/K$ then gives your result.
However, I am not completely sure if the $\alpha_i$ do form an integral basis necessarily. The argument I presented only proves that $\mathcal O_L$ is itself finitely generated and hence finite over $\mathcal O_K$ containing the latter (which is the question asked in the title and what the linked answer is concerned with; please correct me if your questions was something else).