= Solution
The <polynomial> $T^4-2$ is irreducible by <Eisenstein criterion> at two, so $1,\alpha,\alpha^2,\alpha^3$ is a rational basis and each of these elements is integral. The nontrivial automorphism of $K/k$ sends $\alpha$ to $-\alpha$. For an <algebraic integer> $\theta$, its <relative trace> and the trace of the <algebraic integer> $\alpha\theta$ are
$$
\operatorname{Tr}_{K/k}\theta=2a+2c\sqrt2,\qquad
\operatorname{Tr}_{K/k}(\alpha\theta)=4d+2b\sqrt2.
$$
Both are in $\mathcal O_k$, proving $2a,2b,2c,4d\in\mathbb Z$. Multiplying $\theta$ by its conjugate gives the <field norm>
$$
\operatorname N_{K/k}\theta
=(a+c\sqrt2)^2-\sqrt2(b+d\sqrt2)^2
=(a^2+2c^2-4bd)+(2ac-b^2-2d^2)\sqrt2.
$$
Again this is in $\mathcal O_k$, so both displayed coefficients are integers.
To finish the denominator argument, put $A=2a$, $B=2b$, $C=2c$, $D=4d$, all integers. The two norm coefficients imply
$$
A^2+2C^2-2BD\equiv0\pmod4,\qquad
4AC-2B^2-D^2\equiv0\pmod8.
$$
The second congruence modulo two first gives $D$ even. Modulo four it then gives $B$ even. Therefore $2BD$ vanishes modulo four, and the first congruence reads $A^2+2C^2\equiv0\pmod4$. The possible squares modulo four force $A$ even and then $C$ even. Returning to the second congruence now gives $D^2\equiv0\pmod8$, so $D$ is divisible by four. Thus $a,b,c,d$ are all integers.
Conversely, every integer linear combination of these powers is integral. We have proved the <integral basis of the fourth-root-of-two field>:
$$
\boxed{\mathcal O_K=\mathbb Z[\alpha],\qquad1,\alpha,\alpha^2,\alpha^3\text{ is an integral basis}.}
$$
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