Question: $p,q$ primes. 

prove $p=q$ if and only if $p$ divides $q$.

$p|q$ stands for '$p$ divides $q$'

$p|q\Leftrightarrow p=q$

$\Leftarrow$:

$p(1)=q$ and therefore $p|q$

$\Rightarrow$:

if $p=\pm 1$, $p|q\Rightarrow p=q$ does not hold:

all you need is one case to disprove an implication: $1|2 \rightarrow 1=2$, $-1|2 \rightarrow -1=2$ evaluate to $true\rightarrow false$, $true\rightarrow false$ which both evaluate to $false$. 

in turn, $p|q\Leftrightarrow p=q$ does not hold for $p=\pm 1$

for $p\neq \pm 1$: 

proof by contraposition: $[p|q\Rightarrow p=q]\Leftrightarrow [p\neq q\Rightarrow p\nmid q]$

$p\neq q\Rightarrow p\nmid q$:

$p\neq q$ 

$\therefore q=p+r(r$ an integer) $=p+p\_1p\_2...p\_n;\space$ primes $ p\_1,p\_2,...p\_n\neq p$; they cannot equal $p$ because then the prime $q$ would be divisible by something other than itself and $1$, namely $p$.

$\therefore p\nmid q$ 

Please verify proof.

Answer: 
$(\Leftarrow)$ looks good.

For $(\Rightarrow)$ I would say: 

If $p\mid q$ then as $q$ is a prime, $p=\pm1$ or $p=\pm q$ but $p$ is a prime so $p=q$.

I am using: If $q>1$ then $q$ is a prime if and only if the following is true: $d\mid q$ implies $d=\pm1$ or $d=\pm q$.
If $p|q$ and $q|p$, then $p = aq$ and $q = bp$, where $a, b$ are integers. Combining these conditions we get $p = abp$, so $p(1-ab) = 0$. There are no divisors of zero in $\mathbb Z$, so $1 = ab$. The only units (invertible elements) in the ring of integers are $1$ and $-1$, so $p = \pm q$. But you were assuming that both $p, q$ are prime (therefore positive) and the conclusion is that $p = q$.