$\mathcal O_K$ is closed under addition and multiplication

Case $d\equiv2,3\pmod4$ ($\mathcal O_K=\mathbb Z[\sqrt d]$, i.e. all elements $a+b\sqrt d$ with $a,b\in\mathbb Z$): addition is componentwise, $(a+b\sqrt d)+(c+e\sqrt d)=(a+c)+(b+e)\sqrt d$, clearly again of this form. For multiplication, $(a+b\sqrt d)(c+e\sqrt d)=(ac+bed)+(ae+bc)\sqrt d$, and since $a,b,c,e,d\in\mathbb Z$, both $ac+bed$ and $ae+bc$ are integers, so the product stays in $\mathbb Z[\sqrt d]$.

Case $d\equiv1\pmod4$ ($\mathcal O_K=\mathbb Z[\omega]$ with $\omega=\tfrac{1+\sqrt d}2$): first compute $\omega^2$ directly. $\omega^2=\left(\tfrac{1+\sqrt d}2\right)^2=\tfrac{1+2\sqrt d+d}4=\tfrac{1+d}4+\tfrac{\sqrt d}2$. Since $\sqrt d=2\omega-1$, this is $\tfrac{1+d}4+\omega-\tfrac12=\tfrac{d-1}4+\omega$. Because $d\equiv1\pmod4$, $t:=\tfrac{d-1}4$ is an integer, so $\omega^2=t+\omega$: a genuine integer relation showing $\omega$ satisfies the monic integer polynomial $x^2-x-t$, confirming $\omega$ itself is integral (as it must be, matching Theorem 1), and crucially expressing $\omega^2$ again as an integer combination of $1$ and $\omega$.

Now take general elements $a+b\omega$ and $c+e\omega$ with $a,b,c,e\in\mathbb Z$; addition is again componentwise and clearly closed. For the product, $(a+b\omega)(c+e\omega)=ac+(ae+bc)\omega+be\,\omega^2=ac+(ae+bc)\omega+be(t+\omega)=(ac+bet)+(ae+bc+be)\omega$. Both coefficients $ac+bet$ and $ae+bc+be$ are integers (products and sums of integers), so the product lies in $\mathbb Z+\mathbb Z\omega=\mathbb Z[\omega]$.

In both cases, closure under addition and multiplication is verified by direct, elementary computation, confirming $\mathcal O_K$ is genuinely a subring of $K$ containing $\mathbb Z$ in every quadratic field, exactly as the name "ring of integers" promises.