for that question i tried answering it this way,
let b be the infinimum, then for all n in N, b+ 1/n is not a lower bound.
so there exists a sequence xn in A such that
xn < b + 1/n
xn-b < 1/n
{1/n} --> 0, which implies that xn ---> b.
does this make sense or did i do it all wrong?
Formatting aside, it makes sense, and is a sensible way to do things. I think it would be better if you made it clear that you were using the squeeze theorem. You've got this sequence $x_n$ satisfying $$b \le x_n < b + \frac{1}{n}.$$ Since $b$ and $b + \frac{1}{n}$ are sequences that converge to $b$, then so does $x_n$ by the squeeze theorem.
Note also that it's important to state the lower bound (which you didn't in your proof). Without having a lower bound (which comes from the infimum being a lower bound of the set), you cannot conclude convergence. For example, $b - n < b + \frac{1}{n}$ for all $n$, but we cannot conclude that $b - n \to b$.