Is this a valid proof for $\sqrt5$ being irrational? 'Prove by contradiction that $\sqrt5$ is irrational.'
Proof: Assume that $\sqrt5$ is rational i.e. $\sqrt5 = p/q,$ where $p,q \in \Bbb Z$.
Then $\sqrt5q = p$
Now for p to be an integer, $\sqrt5q$ must be an integer, i.e $q=\sqrt5$, $2\sqrt5$, $3\sqrt5$, ...
$\implies$ q must be irrational for p to be an integer.
$\implies$ $p,q \notin \Bbb Z$.
Contradiction. Therefore $\sqrt5$ is irrational. #
 A: Your proof is not correct because you have to prove that $q= k\sqrt{5}$ is irrational for every integer  $k$ by proving that $\sqrt{5}$ is irrational.
Note that if we multiply a non-zero integer and an irrational the product will be an irrational number.(You can prove this for practise)
So you assume what you want to prove.
Here it is a valid proof which i hope it will help you:

Assume that $\sqrt{5}=\frac{m}{n}$ where $(m,n)=1$.
We assume that $(m,n)=1$ because if its not then we can cancel a priori every common factor of the numerator and denominator until we remain with a fraction $\frac{s}{l}$ in its lowest terms namely  $(s,l)=1$.
Thus $$5n^2=m^2 \Rightarrow 5|m^2 \Rightarrow 5|m$$ because $5$ is a prime number.
Also because $5|m$ we have that  $m^2=25s^2$
Thus $$5n^2=25m^2 \Rightarrow n^2=5m^2 \Rightarrow 5|n$$
Now we have a contradiction because $5|n$ and $5|m$ and we assumed that $(n,m)=1$
So $\sqrt{5}$ is an irrational.

A: This is wrong.
Your statement rhat $\sqrt{5}q$ must be an integer does not imply that $q$ is irrational.
A: 
Note that "$\Longrightarrow$ q must be irrational for p to be an integer." is not a correct statement. It includes the assumption that $\sqrt5$ is not a rational number which you want to prove. (See more in the end)

Here is an alternative way:
Let $\gcd(p,q)=1$, 
$\sqrt5p=q\Rightarrow 5p^2=q^2\Rightarrow 6p^2=p^2+q^2$.
Then $3\mid p^2+q^2$, but $p^2+q^2\equiv 1,2(\mod 3)$ since $\gcd(p,q)=1$ and $p^2\equiv 0,1 (\mod 3)$ and $q^2\equiv 0,1 (\mod 3)$, which leads to contradiction.
Added:
"Now for p to be an integer, $\sqrt5 q$  must be an integer, i.e $q=\sqrt5,2\sqrt5,3\sqrt5\dots$" also a wrong statement in a way. 
For example, let $2=\dfrac{p}{q}\Rightarrow q=\dfrac{p}{2}$. Now you you would write...
"Now for q to be an integer, $\dfrac{p}{2}$ must be an integer, i.e $p=\dfrac{1}{2},\dfrac{2}{2},\dfrac{3}{2},\dots$"
But your assumption was $p$ is an integer. But here isn't for each value of $q$. Does it imply $2$ is not a rational number!!!
A: No, your proof is not valid. The mistake happens in two lines:

First mistake:

Now for p to be an integer, $\sqrt5q$ must be an integer, i.e $q=\sqrt5$, $2\sqrt5$, $3\sqrt5$, ...

You didn't list all the possibilities here. For example, $\sqrt{5}q$ is an integer if $q=\frac{9}{\sqrt{5}}$, but you didn't list that possibility.

Second, much bigger mistake:

$\implies$ q must be irrational for p to be an integer.

How do you know this? This is not true for $\sqrt{4}$, so why should it be true for $\sqrt{5}$?
(yes, it is true for $\sqrt 5$, but you have to prove it).
