Is this Galois theory proof of Fundamental Theorem of Algebra correct?
Although there is a lot of correct material in this proof, I see two flaws. One of them is critical, the other is superficial.
Flaw #1: The critical flaw is that you are not postulating an algebraic extension of $\mathbb{C}$. Thus in some sense the proof never begins. The statement that $\mathbb{C}$ is algebraically closed is the statement that if $K$ is an algebraic extension of $\mathbb{C}$, then $K=\mathbb{C}$. Thus you should begin the proof by supposing that $K$ is any algebraic extension of $\mathbb{C}$. This is not how you defined $K$. You introduced it as a Galois extension of $\mathbb{R}$ containing $\mathbb{C}$, which is guaranteed to exist by Dummit&Foote corollary 23. Thus when you prove things about $K$, you are not proving them about any algebraic extension of $\mathbb{C}$ but only about a specific one you have constructed in the proof.
To drive the point home, you don't even need corollary 23 to construct a finite Galois extension $K$ over $\mathbb{R}$ containing $\mathbb{C}$. $K=\mathbb{C}$ is already such an extension. So you could have replaced the sentence "$\mathbb{C}$ is contained in a finite Galois extension $K$ over $\mathbb{R}$" with the sentence "$K=\mathbb{C}$ is a finite Galois extension of $\mathbb{R}$" and the logical work of the sentence wouldn't really have changed. But then if afterwards you proved that $K=\mathbb{C}$, you wouldn't have proved anything at all.
Flaw #2: It is the case that $p$-groups have subgroups of every order dividing the group order (I assume this is what you mean "all orders"), and that $p$-groups have normal subgroups of each of these orders as well, but it is not true that every subgroup of a $p$-group is normal, since nonabelian $p$-groups do exist. What is true is that for each order dividing the group order, there exist subgroups of that order, at at least one of them is normal.
Yes, I believe it is correct, although I haven't read your post in detail - there certainly is a correct proof along these lines! I usually state the two assumptions as
All polynomials of odd degree over the real numbers have at least one root in ${\mathbb R}$.
All positive real numbers have a square root.
They can both be proved the Intermediate Value Theorem, which is a moderately elementary result in calculus. Using 2, you can prove by direct calculation that all complex numbers have a square root.
So this is the most algebraic proof, in that uses the least amount of calcualus/analysis. It is also possible to avoid using Sylow's Theorem, thereby reducing the algenra required.