How do you show that the degree of an irreducible polynomial over the reals is either one or two? The degree of irreducible polynomials over the reals is either one or two.
Is it possible to prove it without using complex numbers? Or without using fundamental theorem of algebra?

1) If you know that every irreducible polynomial over $\mathbb R$ has degree $1$ or $2$, you immediately conclude that $\mathbb C$ is algebraically closed:  
Else there would exist a simple  algebraic extension $\mathbb C\subsetneq  K=\mathbb C(a)$ with $[K/\mathbb C]=\operatorname {deg}_\mathbb C a=d\gt 1$.
Then $K=\mathbb C(a)=\mathbb R(i,a)=\mathbb R(b)$ for some $b\in K$ by the primitive element theorem
But then the minimal polynomial $f(X)\in \mathbb R[X]$ of $b$ over $\mathbb R$  would be  irreducible  over $\mathbb R$ and have degree $\operatorname {deg} f(X)=2d\gt 2$, a contradiction to our hypothesis. 
2) That said it is possible to prove  that every irreducible polynomial over $\mathbb R$ has degree $1$ or $2$ without using the Fundamental Theorem of Algebra for $\mathbb C$.
The method is due to Lagrange and is described in Samuel's Algebraic Theory of Numbers, pages 44-45.
The method consists in inducting on the largest power $r$ of $2$ dividing the degree $d=2^rl$ ($l$ odd) of an irreducible real polynomial,  the result being clear  for $r=0$ i.e. for odd $n$.
The proof (highly non trivial) proceeds by a clever application of Viète's formulas expressing the coefficients of a polynomial as symmetric functions of the roots of that polynomial.   
3) Another real methods  proof uses Galois theory and Sylow $2$-groups.
It can be found in Fine-Rosenberg's Theorem 7.6.1
That elementary and pedagogical book is entirely devoted to all kinds of  proofs of the Fundamental Theorem of Algebra.  
