Cube root of $4$ is irrational I would like to see if this proof is correct. I will really appreciate it.  

Suppose that ${4}^{\frac{1}{3}}$ is rational. Then there exists $a,b
 \in \mathbb{Z}$ such that $\mathrm{gcd}(a,b) = 1$ with $\frac{a}{b} =
 4^{\frac{1}{3}}$.  
Then $a^3 = 4b^3$. (So from here I want to show that $a,b$ are in fact
  not coprime).  
Since from assumption, $a,b$ are coprime, then $a^3 | 4$. This implies
  $a$ is even (I think?!).  
So $a = 2k$ for some $k\in \mathbb{Z}$.  
Hence, $4b^3 = 8k^3 \implies b^3 \stackrel{(\star)}{=} 2k^3.$ Since $b
> \nmid a$ then $b\nmid 2k$ so $b \nmid 2$ and $b\nmid k$ by Euclid's
  Lemma.   But if this is true, then $b^3 \nmid 2$ and $b^3 \nmid k^3$.
  So $b^3 \neq 2k^3$ which contradicts $(\star)$.

This feels weird, because I feel like I should contradict one of my hypotheses, not something I found along the way...
 A: In your proof $a$ might be $1$.
If $a=1$ then $b$ must be a prositive integer because $\sqrt[3]{4}$ is a positive number,i.e $b \in \mathbb{N}$
So you can say that if $a=1$ then from the fact that $b \in \mathbb{N}$ we have that $1=b \sqrt[3]{4}>1$ which is a contradiction.
Now you can prove your statement in this simpler way if you want:

$2$ is a prime number therefore if $2|a^k$ for some positive integer $k>1$ then $2|a$
We have that $4|a^3$ thus $2|a^3 \Rightarrow 2|a \Rightarrow a=2k$ 
From this we see that  $$4b^3=8k^3 \Rightarrow b^3=4k^3 \Rightarrow 2|b^3 \Rightarrow 2|b$$
Thus we have : $$2|a$$ $$2|b$$
So from this we contradict our assumption about $(a,b)=1$
and the proof is finished.

A: proof-verification:

Suppose that ${4}^{\frac{1}{3}}$ is rational. Then there exists $a,b \in \mathbb{Z}$ such that $\mathrm{gcd}(a,b) = 1$ with $\frac{a}{b} = 4^{\frac{1}{3}}$.  
Then $a^3 = 4b^3$. (So from here I want to show that $a,b$ are in fact not coprime (so that I would have a contradiction)).  
Since from the assumption, $a,b$ are coprime, then $a^3 | 4$. (Why? Maybe you want to conclude that $4|a^3$?) This implies $a$ is even (I think?!). (Why? This step is unclear.) 
So $a = 2k$ for some $k\in \mathbb{Z}$. (Assuming you have shown $a$ is even, I keep reading.)
Hence, $4b^3 = 8k^3 \implies b^3 \stackrel{(\star)}{=} 2k^3.$ Since $b \nmid a$ then, i.e. $b\nmid 2k$, so $b \nmid 2$ and $b\nmid k$ by Euclid's Lemma. (Euclid's lemma says: "If a prime p divides the product ab of two integers a and b, p must divide at least one of those integers a and b." I don't see how it is used here. It seems that you use it in a wrong way. Your argument breaks down and I would stop reading from here.)

A: It's not clear how $a^3\mid 4$, how $a$ is even, nor how $b^3\neq 2k^3$; besides: $a^3\mid 4$ could imply $a^3=1$ a priori - you'd have to argue that $a\neq 1$ by approximating $4^\frac13$.
Since $a^3=4b^3$, we have $4\mid a^3$, which implies that $a$ is even as $2$ is prime.
If you deduce any contradiction from the hypothesis that $4^{\frac13}$ is rational, it means that it is indeed irrational. There's nothing weird about it.
A: You can make it stronger and quicker.
But first we need to be clear:
Lemma:  If $a$ and $b$ are coprime then $a|kb \implies a|k$.
Lemma 2: If $a$ and $b$ are coprime then $a^m|kb^n \implies a^m|k$.
These lemmas are acceptable if we assume every number has a  prime factorization.
So $a^3 = 4b^3$ would imply $a^3|4$.
But that is a REALLY strong statement! 
$a^3 = 4b^3 \implies 1 = \frac {4}{a^3}b^3$ but $1$ is unitary so $\frac 4{a^3} =1$ and $a^3 = 4$.  So the cube root of $4$ must be an integer if it is rational!
Well.  $a \le 0\implies a^3 \le 0 < 4$.  And $1^3 = 1 \ne 4$.  And if $a \ge 2$ then $a^3 \ge 2*a^2 \ge 2*2*a \ge 2*2*2 = 8 > 4$.
.... or ... if we can assume that every integer has a unique prime factorization.  (Cany you in your class and these stage?) then $a^3 = \prod p_i^{3a_i} = 2^2$ is .... impossible.
