I thought I proved something until I looked at the hints, which leads me to believe my proof attempt is too simple.
$\phi:\mathbb{R}\rightarrow \mathbb{R}$ is a unital ring isomorphism (i.e. $\phi(1)=1$). I want to prove that $\forall x\in\mathbb{R}$, $\phi(x)=x$.
I tried this: since $\phi$ is a homomorphism, $\phi(x+y)=\phi(x)+\phi(y)$. So $\phi(x)=\phi(1+1+...+1)=\phi(1)+\phi(1)+...+\phi(1)$ (with $x$ number of $1$'s). Then $\phi(x)=x1=x$. The hints for the proof say to use the fact that $\forall a,b\not\in\mathbb{Q} a<b$ $\exists c \in \mathbb{Q}$, and similarly for between rationals. It also mentioned the use of square roots. First question, why is my attempt flawed, and secondly, how might I use these instructions to complete the proof?
Your attempt is correct but it does not applies to all $x\in \mathbb{R}$. For example, what if $x=\sqrt{2}$. You can't say that $\sqrt{2}=1+1+...+1$ taking $1's$ $\sqrt{2}$ times. One strategy is to show that $\phi(x)=x$ $\forall x\in \mathbb{Q}$ (your attempt works here) and to observe that $\phi$ is continuous thus extending $\phi$ to all of $\mathbb{R}$.
Edit :- Let $x,y\in \mathbb{R}$ with $|x-y|<\frac 1n$ then $|\phi(x)-\phi(y)|<\frac 1n\phi(1)$ (since if $x\geq 0$ then $\phi(x)\geq 0$)...hence $\phi$ is continuous.