Prove whether a relation is an equivalence relation Define a relation $R$ on $\mathbb{Z}$ by $R = \{(a,b)|a≤b+2\}$.
(a) Prove or disprove: $R$ is reflexive. 
(b) Prove or disprove: $R$ is symmetric.
(c) Prove or disprove: $R$ is transitive.
For (a), I know that $R$ is reflexive because if you substitute $\alpha$ into the $a$ and $b$ of the problem, it is very clear that $\alpha \leq \alpha + 2$ for all integers.
For (b), I used a specific counterexample; for $\alpha,\beta$ in the integers, if you select $\alpha = 1$, and $\beta = 50$, it is clear that although $\alpha ≤ \beta + 2$, $\beta$ is certainly not less than $ \alpha + 2$.
However, for (c), I am not sure whether the following proof is fallacious or not:
Proof: Assume $a R b$ and $b R g$;
Hence $a ≤ b + 2$ and $b ≤ g + 2$
Therefore $a - 2 ≤ b$ and $b ≤ g + 2$
So $a-2 ≤ b ≤ g + 2$
and clearly $a-2 ≤ g+2$
So then $a ≤ g+4$
We can see that although $a$ might be less than $ g+2$,
it is not always true. 
Therefore we know that the relation $R$ is not transitive.
QED.
It feels wrong

You are right, the attempt to prove transitivity doesn't work. But your calculation should point towards a counterexample. 
Make $a \le b+2$ in an extreme way, by letting $b=a-2$. Also, make $b\le g+2$ in the same extreme way. Then $a \le g+2$ will fail.  Perhaps work with an explicit $a$, like $47$.
