In his doctoral thesis, Gauss gave a proof of fundamental theorem of algebra for real polynomials, based on geometric arguents. Later in his life he expanded the proof to complex polynomials. A nice account can be found here.
The problem with Gauss's proof was that he assumed some topological arguments, which were not proven rigorously till a century later. A proof is presented here which starts in Gauss's lines, then replaces his topological arguments with least upper bound for real numbers and continuity considerations only. Gauss was aware of the least upper bound property (LUB) for real numbers (in fact he co-discovered it). So his topological arguments could have been replaced by the following method, which is why I am calling it "Correction". Any help with verification will be greatly appreciated.
Theorem: Every complex polynomial has a complex root.
Background: We only consider polynomials which are not factorizable into lower degree polynomials of degree >1. For factorizable polynomials, the proof can be applied on the factors.
Complex numbers are assumed to exists and it is known that they form a field (So that basic algebraic operations and distributive law are well defined). Modulus function is also defined.
Gauss's proof:
Rewrite $z=re^{i\theta}$. Then $z^k=r^{k} (cos(k\theta) +i sin(k\theta))$
Then $\sum_{k=0}^{n}c_{k}z^{k}=\sum_{k=0}^{n} | c_{k} | r^{k}(cos(k\theta+\phi_{n}) +i sin(k\theta+\phi_{k}))$
So real part of $R(r,\theta)=\sum_{k=0}^{n} | c_{k} | r^{k} cos(k\theta+\phi_{k} )$ and the imaginary part is $I(r,\theta)=\sum_{k=0}^{n} | c_{k} | r^{k} sin(k\theta+\phi_{k})$.
Taking $c_{n}=1$, Gauss showed that for sufficiently large $r=L$, $R(L,\theta)$ changes sign in the interval $\theta \in (\frac{\pi k'}{n}-cos^{-1}\epsilon,\frac{\pi k'}{n}+cos^{-1}\epsilon), \quad k=0,1,2,...,2(n-1)$. Applying intermediate value theorem , this means that $R(L,\theta)$ has roots in these intervals of $\theta$, for any $r$ large enough.
On the other hand, for L sufficiently large, $I(L,\theta)$ changes sign in the interval $\theta \in (\frac{\pi}{2n}+ \frac{\pi k'}{n}-cos^{-1}\epsilon,\frac{\pi}{2n}+\frac{\pi k'}{n}+cos^{-1}\epsilon)$ and thus has a root in every interval.
If $\epsilon>0$ is taken sufficiently small, al these intervals have zero overlap. So on the circle $r=L$, we have $2n$ roots of $R$ and $I$ interlacing. All roots are exhausted. Gauss then deduced the existence of a root based on this fact and the following assumption:
Lemma 1: If two continuous bivariate function have interlacing roots on a closed curve, they must have a simultaneous roots inside the curve.
The proof of this is quite hard, based on topological arguments (connected curves, Jordan curve theorem etc). Gauss speculated that this could be proven (his speculations were often correct and groundbreaking) but he did not have a proof.
My proof:
Using the same $R$ and $I$ as Gauss, only least upper bound and continuity arguments can be used to prove FTA. $R$ and $I$ can also be written in terms of $(x,y),z=x+iy$ in the following way.
$$\left[\begin{array}{l}R(x,y)\\I(x,y)\end{array}\right]= \sum_{k=0}^{n} \left[\begin{array}{l}Re(c_{k})&-Im(c_{k})\\Im(c_{k})&Re(c_{k})\end{array}\right] \left[\begin{array}{l}x&-y\\y&x\end{array}\right]^{k} \left[\begin{array}{l}1\\0\end{array}\right]$$
If we take $\bar{y}$ large enough, $r$ is also large enough for any $x,\bar{y}$. For all
the disjoint $\theta$ intervals Gauss created in this proof, $x=\bar{y} tan(\theta)$ also generate disjoint intervals. Usings intermediate value theorem on $x$, it is starigtforward to deduce that $\exists y_{1}<0$, such that $R(x,\bar{y})$ and $I(x,\bar{y})$ have interlacing roots $\forall \ \bar{y}<y_{1}$. Looking at the matrix formulation of $R$ and $I$, it can be seen directly that the maximum power of $x$ in $R(x,\bar{y})$ and $I(x,\bar{y})$ are $n$ and $n-1$ respectively. So all real roots exist and interlace $\forall \ \bar{y}<y_{1}$.
Now comes LUB based 'Correction':
1. $R(x,0)$ does not have any real root. Wrong, see correction at {*} in the end of this article
2. $\exists y_{1}<0$, such that $R(x,\bar{y})$ and $I(x,\bar{y})$ have maximum possible number of real roots, which interlace $\forall \ \bar{y}<y_{1}$
3. LUB applied on 1. and 2. says that $\exists y_{sup}<0$ such that both $R(x,\bar{y})$ and $I(x,\bar{y})$ have maximum number of real roots $\forall \bar{y}<y_{sup}$ and at least one of them has less number of real roots for $y_{sup}<\bar{y}<y_{sup}+\delta$, for some $\delta >0$.
4. Using intermediate value theorem and due to the finite number of roots, it can be shown that the roots of both $R(x,\bar{y})$ and $I(x,\bar{y})$ vary continuously with $\bar{y}, \quad \forall \bar{y}<y_{sup}$. Question is what happens at $\bar{y}=y_{sup}$.
5. $R(x,\bar{y})$ and $I(x,\bar{y})$ are both bivariate polynomials of $x$ and $y$. Thus the roots on x are clearly bounded $\forall \bar{y}<y_{sup}<0$. Thus $liminf_{\bar{y} \rightarrow y_{sup}}$ and $limsup_{\bar{y} \rightarrow y_{sup}}$ of the roots exist for any indexed root. If the $limsup$ and $liminf$ are different for any indexed root, $R=0$ or $I=0$ crosses any value between these two infinite times. For fixed $x$, $R$ or $I$ are polynomials in $y$ and cannot have infinite number of roots. So the $liminf=limsup$ (implying that the roots approach limits as $\bar{y} \rightarrow y_{sup}$). Thus $R(x,y_{sup})$ and $I(x,y_{sup})$ have maximum possible real roots.
At least one of the roots do not exist for slightly large $\bar{y}$. If all roots of $R(x,y_{sup})$ and $I(x,y_{sup})$ are distinct, intermediate value theorem can be applied on the larger side of $\bar{y}$ to have a contradiction. Thus the only possibility is that there is a root on $x$ which has same sign on both sides on the line $\bar{y}=y_{sup}$. This is clearly a root with multiplicity, implying that two of the root lines meet. It is easy to show that two consecutive roots meet at $\bar{y}=y_{sup}$
Without loss of generality, let us say two consecutive roots of $R$ meet at $\bar{y}=y_{sup}$. $\forall \bar{y}<y_{1}$, there is a root of $I$ between these two roots. Again, straighforward application of LUB gives that $\exists y_{sup}<y<y_{1}$ and the corresponding root of $I$ mentioned in the last line such that $R(x,y)=I(x,y)$. This proves FTA.
I did not want this to be a very long post. So Sorry for being terse in the derivations.
{*}Update 24 April 2016:
In my 'Correction', point 1 only works for real polynomials. Slightly more work needs to be done for complex polynomials. If either the real or complex part of the polynomial ($R$ and $I$)has less than $n$ or $n-1$ roots at $\bar{y}=0$, the rest or the arguments go through. If not, since maximum possible real roots exists for $R(x,\bar{y})$ and $I(x,\bar{y})$ for all $\bar{y}$, the roots vary continuity according to point 4. Then following possibilities exists
Case 1: For some $\bar{y}$, the roots of $R(x,\bar{y})$ and $I(x,\bar{y})$ do not interlace. LUB and continuity arguments can be directly applied on this to deduce the existence of a simultaneous solution of $R(x,\bar{y})$ and $I(x,\bar{y})$.
Case 2: The interlacing is always preserved. Writing $R(x,\bar{y})$ as a function of $x$ and applying the Fujiwara bound for the upper bound of the roots, it can be seen that there cannot exist a roots of $R(x,\bar{y})$ for large enough $x$ in the region $\bar{y}<mx$, for some $m\in \mathbb{R}$. However, this is impossible since rewriting $R$ as $R(r,\theta)$ gives that for small, say $\theta^{*}=tan^{-1}\frac{m}{2}$, $\exists \bar{r}$ and $\theta(r)$ such that $R(r>\bar{r},\theta(r)))=0$ satisfying $|\theta(r)|<|\theta^{*}|$. If anyone follows up in details to this point and wishes to have a full explanation of this, I will be glad to provide one.