If $u$ is a unit in $S$, then $u$ is a unit in $R$ I think the proof is fairly simple...but of course one of my big problems in math is overdoing simple problems. So..

Let $R$ be a ring with identity, and let $S<R$ be a subring containing the identity.  Prove that if $u$ is a unit in $S$, then $u$ is a unit in $R$.  Show by example that the converse is false.

So, let $u$ be a unit in $S$.  By definition of a unit, there exists a $v\in S$ such that $1=uv=vu$.  However, since $1\in R$, then by closure, $uv\in R$, and thus, $u$ is a unit in $R$.
As far as the converse, take the ring of Gaussian integers $\mathbb{Z}[i]=R$, and $S=\mathbb{Z}$.  If the converse was true, this would say that $i=
\sqrt{-1}\in S$, which is clearly false, since $1=-i\cdot i$.
How is this?

For the first part: That’s not quite correct. You need to argue that $v ∈ R$ rather than $1 = uv ∈ R$.
For the counterexample: I think you should rather disprove that

Whenever $S ⊂ R$ is a subring of a ring $R$, and $u ∈ S$ is a unit in $R$ (that is there is some $v ∈ R$ with $uv = 1$ in $R$) then $u$ is also a unit in $S$ (that is there is some $v' ∈ S$ with $uv' = 1$ in $S$).

Your example doesn’t work in that case. Rather think of $ℤ ⊂ ℚ$.
