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My question is about the following lemma (where the Extreme Value Theorem is assumed);

$(1)$Let $f : mathbb{C} → mathbb{C}$ be any polynomial function. Then there exists a point $z_0$ $∈ mathbb{C}$ where the function $|f|$ attains its minimum value in $mathbb{R}$.

$Proof.$ If $f$ is a constant polynomial function, then the statement of the Lemma is trivially
true since $|f|$ attains its minimum value at every point in $mathbb{C}$. So choose, e.g., $(z_0) = 0$.
If $f$ is not constant, then the degree of the polynomial defining $f$ is at least one. In this
case, we can denote $f$ explicitly as in Equation (3.1). That is, we set

$f(z) = a_n$ $z^n + · · · + a_1z + a_0$

with $a_n≠0$. Now, assume $z ≠ 0$, and set $A = max${|$a_0$|, . . . , |$a_{n−1}$|}$.$ We can obtain a lower
bound for $|f(z)|$ as follows:

$|f(z)| = |a_n|$ $|z|^n$ $|1 + ((a_{n-1}/a_n)(1/z))$ $+...+$ $((a_0/a_n) (1/z^n))|$

I'm missing the second part of the inequality because I don't know how to format it, but if anyone would like to do so for me you can find it on page 34 of this link,
https://www.math.ucdavis.edu/~anne/linear_algebra/mat67_course_notes.pdf

My question, however, is about why one can obtain a lower bound like this, why does it work and how does it help us to prove statement $(1)$.

I'd appreciate secondary sources for further reading as well, and for a bonus explain why this relates to linear algebra.

The idea behind the proof is that a continuos function on a compact set is bounded and attains its bounds.

In your case you choose a sufficiently large R such that for $|z|>R$ the terms $|frac{a_i}{z^n}|$ are negligible and therefore the value of $f(z)$ is approximately equal to $a_nR^n$. Moreover, on the compact set $|z| leq R$, p is continuous and therefore bounded below by some M >0 and moreover this is attained. For R large we can ensure $R^n|a_n| > M$ and therefore M will be a global minimum and it will also be attained.