Proof read from "A problem seminar" May you help me judging the correctness of my proof?:
Show that the if $a$ and $b$ are positive integers, then
$$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n$$
is integer for only finintely many positive integers $n$
We want $n$ so that
$$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n\equiv0\pmod{1}$$
So we know by the binomial theorem that
$(an+b)^k\equiv b^n\pmod{n}$ for positive $k$
Then,
$$\left(a+\frac{1}{2}\right)^n\equiv(1/2)^n\pmod{1}$$
and similarly with the $b$
So 
$$\left(a+\frac{1}{2}\right)^n+\left(b+\frac{1}{2}\right)^n\equiv 2*(1/2)^n\pmod{1}$$
Therefore, we want $2*(1/2)^n$ to be integer, so that $2^n|2$
clearly, the only positive option is $n=1$
(Editing, my question got prematurely posted. Done)

Andre gave you a solution, but you wanted proofreading, so I'll point out two mistakes you've made (I'm not saying there aren't more :-)).
First, modulo arithmetic assumes you're working with integers, so it cannot be applied on rationals.
Second, when getting from
$$(an+b)^n\equiv b^n\pmod{n}$$
to
$$\left(a+\frac{1}{2}\right)^n\equiv(1/2)^n\pmod{1}$$
You put $n = 1$, but just in some spots (but left it unchanged in the exponents). You cannot use formulas like that. It's like going from
$$n = n$$
and then substituting the first $n$ with $1$, getting
$$n = 1.$$
Of course you'll get $n = 1$ when you use it like that.
