We're given the polynomial $x^{2}-2$ , and we need to prove that it's irreducible in $\mathbb Q$ but reducible in $\mathbb R$.
Writing the polynomial as $(x^{2}-2) = 1.(x^{2}-2)$ ,
$(x^{2}-2)$ can't be a unit in $Q[x]$ whereas $1$ is a unit in $Q[x]$ , hence the given polynomial is irreducible in $Q$.
Can't we apply the same argument for $R$ also ? Since rationals are contained in reals , won't that imply that it's irreducible in $R$ too ?
But we have to prove that it's reducible in $R$ , am I missing something ?
Like the comments said, $x^2-1 = (x-1)(x+1)$, which is reducible in $\mathbb{Q}[x]$, clearly. So, I am going to assume that you mean $x^2+1$, which is not reducible in $\mathbb{R}$, since $\text{discriminate}(x^2+1) = 0^2 - 4(1)(1) = -4 < 0$.
But, to show that it is not reducible in $\mathbb{Q}[x]$, suppose that it is, then it is the product of two degree one polynomials in $\mathbb{Q}[x]$; I.e. $$x^2+ 0x +1 = (x - a)(x -b) =x^2 -(a+b)x + ab $$
Implying $a+b = 0$, and $ab = 1$. Further, $a = -b$ implying $ab = (-b)(b) = 1$, a contradiction.
EDIT: Note, similar logic holds in showing that $x^2-2$ is irreducible in $\mathbb{Q}$. To show that it is reducible in $\mathbb{R}$ it is enough to show that $x^2-2 =(x-\sqrt{2})(x +\sqrt{2})$, noting that $\sqrt{2} \in \mathbb{R}$ and that $0 \neq \text{deg}(x \pm \sqrt{2}) = 1 < \text{deg}(x^2-2) = 2.$