Proof that $x^4+1$ is irreducible over $\mathbb{Q}$

For this polynomial, it can simplified, since we know its factorisation over $\mathbf R$: $$x^4+1=(x^2+1)^2-2x^2=(x^2-\sqrt2x+1)(x^2+\sqrt2x+1),$$ and this factorisation does not have rational coefficients. So by the uniqueness of factorisation, it cannot have another factorisation over $\mathbf Q$.

Your approach in principle works fine you just tried to take a short-cut that is not available, yet is not really needed. Instead:

$$(x^4+1) = (x^2+ax+c)(x^2+bx+d)$$

We get

$$x^4+(a+b)x^3+(ab+c+d)x^2+(ad+bc)x+cd.$$

You get $cd=1$ and thus $c=d^{-1}$. You still get $a=-b$ (from the third power), and then $ad +bc$ becomes $ad-ac$ and further $a(c^{-1}-c)$.

Thus, $a=0$ or $c= c^{-1}$ (that is, $c = \pm 1$).

In the former case, you get $c+ c^{-1}= c+d = 0$ which is impossible. In the latter case, you already dealt with $c=d=1$ and $c= d= -1$ is essentially the same, it yields the impossible $a^2 = -2$.

That said, the Eisenstein approach mentioned in a comment is more elegant, if you have it available.

Your proof is almost fine - you just have to write more generally $$ x^4+1=(x^2+ax+b)(x^2+cx+d), $$ where we obtain of course $bd=1$, which over the integers only leaves the possibilities $b=d=1$, or $b=d=-1$. In both cases, the associated Diophantine equations have no solution, and this is easy to see (we obtain $c=−a$, and $a^2 =b+d=±2$, which is not possible).

Furthermore, the polynomial $x^4+1$ is the $8$-th cyclotomic polynomial, and hence irreducible by the theorem of your algebra lecture. Of course, there are several other methods to prove this directly, e.g., writing $f(x)=x^4+1$, and applying Eisenstein to $$ f(x+1)=x^4+4x^3+6x^2+4x+2. $$