It is indeed interesting that such an elementary 'high-school level' proof exists; the OP might be interested in looking at another "in the same vein" question from this site:
Proof of fundamental theorem of algebra in Baby Rudin
The best way to answer your questions is to present a high school level demonstration (including some motivational and explanatory text) of the FTA using elementary ideas such as Lemma 1 and Lemma 2.
Let $f(z)=\sum_{k=0}^na_kz^{k}$ be a polynomial defined on $\mathbb {C}$ with
all coefficients $a_k \in \mathbb {C}$ and $a_n \ne 0$
(we also assume that $n \ge 1$).
Proposition 1: $\lim_{z\to\infty} f(z) = \infty$.
As the complex number $z$ gets farther and farther away from $0$, the highest degree term $a_n z^n$ will begin to 'dominate the action', so that $f(z)$ also gets farther and farther away from $0$.
It isn't easy to crank out a graph of these functions (uh, you would need 4-dimensional graph paper!), but at least we don't have to worry about checking all those $+\infty$ and $+\infty$ cases as we did when studying real-valued functions.
It is worth pointing out that the extended complex numbers, $\mathbb {C}_{\infty} = \mathbb {C} \cup \infty$, can be looked at as a 2-dim sphere, and in higher mathematical discourse it is called the Riemann Sphere. The number $0$ is at the 'south pole' and $\infty$ is at the 'north pole'.
The student already knows how the function $g(z) = z^2$ maps circles $|z| = R$ on the complex plane. A fascinating thing to construct in the mind is to 'see' how this function is mapping $\mathbb {C}_{\infty}$ to itself. This function is a 2:1 surjective mapping, except at the 'poles', $g(0) = 0$ and $g(\infty) = \infty$.
Notice that Proposition 1 can be restated as follows:
Proposition 2: $\lim_{z\to\infty} |f(z)| = +\infty$, and the function
$z \mapsto |f(z)|$
goes from $\mathbb {C}$ to the interval $[0, +\infty)$, or, if you prefer,
$\mathbb {C}_{\infty} \to [0, +\infty]$.
The Extreme Value Theorem also works in this setting, allowing us to state that a minimum 'distance' from $0$ is in the image of $f$:
Proposition 3: There exists a number $c \in \mathbb {C}$ such that
$\tag 1 |f(c)| \le |f(z)|$ for all $z \in \mathbb {C}$.
If we can show that $|f(c)| = 0$, i.e. $f(c) = 0$ we would know that every polynomial of degree greater than $1$ has at least one complex number root. But then you could divide out by $z - c$, and continue this work to get all the roots (of course some roots may come with a multiplicity).
With some fancy footwork you might find a technique showing that $c$ is a root, but it is easier to just go for the contradiction.
Proposition 4: Let $g(z)$ be any non-constant polynomial such that $g(z_0) \ne 0$. Then a complex number $z_1$, close to $z_0$, can be found such that
$\tag 2 |g(z_1)| \lt |g(z_0)|$
Proof
Using the linear variable substitution, $h(z) = g(z + z_0)$, we would have $h(0) = g(z_0)$. So, without loss of generality, we can just assume that $z_0 = 0$. Now when we went to $\infty$ the highest power 'dominates' the action, but as we approach $0$, the term right after the constant is the most important since it goes to $0$ more slowly. By approaching $0$ in the right manner, aligning with the negative of $g(0)$ , we can erase part of the constant 0-th power term. Instead of carefully completing this proof, we leave it to the student; they will have to use the triangle inequality and understand the following key idea:
Lemma: Let $h(z) = a + b z^m$, so that $h(0) = a$. There exist numbers $z_1$ as close to $0$ as necessary satifying
$|h(z_1)| \lt |h(0)| = a$.
Proof
For $0 \lt \lambda \lt 1$, we want to find a $z_1$ with
$h(z_1) = a + b {z_1}^m = \lambda a$. But we can solve for $z_1$,
$\quad z_1 = \sqrt [m] \frac{(\lambda - 1) a}{b}$
Note that $|z_1| = \sqrt [m] {\lambda - 1} \; \sqrt [m] \frac{| a|}{|b|}$, and so $\lambda$ controls the modulus of $z_1$.
Theorem 5: Any non-constant polynomial $f$ over the complex numbers has $n$ roots.
Proof
By Proposition 3 we can find a complex number such that the image under the function $f$ is at a minimum distance from $0$. Assume this distance in not equal to zero. But then by Proposition 4 we can find values in the range of $f$ that are even closer to $0$; this is a contradiction.
