Has the Fundamental Theorem of Algebra been proved using just fixed point theory? Question:

Is there already in the literature a proof of the fundamental theorem of algebra as a consequence of Brouwer's fixed point theorem? 

N.B. The original post contained superfluous information, but it did generate one answer with a source that claims such a proof is impossible, and another answer with a source that claims to carry out precisely such a proof. Clearly these cannot both be correct.
 A: Yes you can prove FTA using BFPT.  See also recent paper by Daniel Reem relating to Open Mapping Theorem.
I often looked at the 1949 note of B.H. Arnold (a professor of mine in bygone times) but missed the flaw.  Maybe someone could post the essence of the retraction by him and Ivan Niven.  Thank you.
You have proved FTA when you prove that any irreducible *real polynomial has degree 1 or two.
The existence of an *irreducible one of higher degree leads to a real vector space of that same dimension on which polynomial product induces a ring structure, in fact, integral domain.
Consider the squaring map on this real algebra, and scale it to map the whole thing to a Cartesian sphere of dimension one less.  Scaling the domain vector variable also, we get a squaring mapping from unit sphere to unit sphere, in fact from RP(n-1) to the sphere. The mapping is topologically continuous (on Hausdorff spaces) due to bilinearity of the original product operation.
The integrity condition (entire domain) shows that this mapping is injective. Everything in sight is compact Hausdorff, so such a 1-1 mapping induces a homeomorphism to the image.
Throw in "connectedness" and the (Brouwer) Invariance of Domain Theorem shows that in fact the image of RP(n-1) must be the whole sphere.  The fact that these two spaces are thus homeomorphic leads to the conclusion that n<=2 .
To show that RP is not topologically the same as S(n-1), construct a non-trivial loop in RP by projecting a longitude from North to South pole on its universal covering sphere.
This loop could not be homotopic to the constant loop (at the chosen base point) since the homotopy would lift to the UC keeping end-points invariant.
What does this have to do with BFPT?  As Dr. Terence Tao points out in his blog, the latter theorem is necessary in some form to prove Invariance of Domain.
A: With regard to the answer already provided: 
The Arnold proof is well known to be erroneous, but a correct (as far as I know) version is cited in an earlier MO post here. In particular, it is a proof of the FTA via the Brouwer Fixed Point Theorem.
The latter source is: 

Some Properties of Continuous Functions. M. K. Fort, Jr. The American Mathematical Monthly, Vol. 59, No. 6 (Jun. - Jul., 1952), pp. 372-375. http://www.jstor.org/stable/2306806.

Edit 1: Todd Trimble has kindly provided a link to the Fort paper that does not require JSTOR access.
Separately, I see the following quotation:
"Recently, there have been very interesting proofs of the Brouwer theorem. Kulpa deduced a generalization of the Brouwer theorem from the Fubini theorem and the Weierstrass approximation theorem, and applied it to give a simple proof of the fundamental theorem of algebra."
The source of this excerpt is: 

Park, S. (1999). Ninety years of the Brouwer fixed point theorem. Vietnam Journal of Mathematics, 27(3), 187-222. http://www.math.ac.vn/publications/vjm/vjm_27/No.3/187-222_Park.PDF.

And the reference under discussion is:

W. Kulpa, An integral criterion for coincidence property, Radovi Mat.6 (1990) 313-321.

I gathered this information at the request of D. Goroff some time ago, at which point my search for the Kulpa paper was, unfortunately, fruitless. 
Edit 2: Karol Szumiło remarks that a friend in one of Warsaw's libraries was able to track down the Kulpa paper! Given the difficulty of finding it, I have uploaded a copy here.
A: Suppose $K/\mathbb C$ is a finite extension of fields. From the multiplication of $K$ we can turn the projectivization $P(K)$ of $K$ as a vector space into a group, which is then a compact connected Lie group. If $g\in P(K)$ is any point, then the map $L_g$ given by left multiplication by $g$ on $P(K)$ is homotopic to the identity map. Its Lefschetz number $\Lambda(L_g)$ is therefore equal to the Euler characteristic of $P(K)$ which is non-zero (because we know that $P(K)$ is a complex projective space!) It follows from Lefschetz's fixed point theorem, the map $L_g$ has a fixed point on $P(K)$. This is only possible if $g$ is the identity element of $P(K)$, and therefore $P(K)$ has exactly one point. 
This means that the extension $K/\mathbb C$ is trivial, and that $\mathbb C$ is algebraically closed.
