Please critique my proof regarding continuity My analysis TA assigned us a proof as practice, but his solution is quite different from mine and I wish to confirm that my logic is sound.
Proposition: Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuous function. Let $(f(x) = 1)~ \forall~ x ~\in~ \mathbb{Q}$. Then $(f(x) = 1)~ \forall~ x \in ~ \mathbb{R}$
Proof: 
Suppose $f(x) \neq 1$ for some $x_0\in\mathbb{R}$. Because $f$ is continuous, $lim_{x\rightarrow x_0} f = f(x_0)$. Let $f(x_0)=A$
Fix $\epsilon > |A-1|$. 
$\exists \delta>0$ st $|x-x_0|<\delta \Rightarrow |A-f(x)| < \epsilon$. This implies that $x \notin \mathbb{Q}$, because if $x \in \mathbb{Q}$ then $|A-f(x)|=|A-1|<\epsilon$ by assumption. But $\mathbb{Q}$ is dense in $\mathbb{R}$, so every $B_{\delta}(x) \cap \mathbb{Q}$ is nonempty. This is a contraction, so $(f(x) =1)~ \forall{x}$.
Thanks.
 A: Your proof is essentially correct, but I believe you could phrase it a little better.

Suppose $f(x) \neq 1$ for some $x_0\in\mathbb{R}$.

Small typo: This should be $x_0$ instead of $x$. Also, you should establish $A$ here already. I'd write:

Let $x_0\in\mathbb{R}$. Suppose $ A := f(x_0) \neq 1$.

Now for another typo:

Because $f$ is continuous, $lim_{x\rightarrow x_0} f = f(x_0)$.

This should be $\lim_{x\to x_0} f(x) = f(x_0)$. (Also observe the different Latex formatting.)

Fix $\epsilon > |A-1|$.

Observe that $\epsilon$ is not a small quantity here, but a big one. I'd call it $M$.

$\exists \delta>0$ st $|x-x_0|<\delta \Rightarrow |A-f(x)| < \epsilon$.

What is $x$? And personally, I like to avoid writing in symbols. It's really hard to read and looks weird, but of course, opinions differ. You definitely have to talk about $x$ though. I'd write

Choose $\delta>0$ such that for all $x \in \mathbb{R}$, $|x-x_0|<\delta $ implies $|A-f(x)| < \epsilon$.

The rest of the proof is fine. Good job! There is a simpler way, though. You don't need contradiction:
Let $y \in \mathbb{R}$. Choose a sequence of rational numbers $(x_n)_{n\in\mathbb{N}}$ such that $x_n \to y$. This is possible by the definition of the reals, or by the fact that $\mathbb{Q}$ is dense in $ \mathbb{R}$. Then we have
$$f(y) = \lim_{n\to\infty} f(x_n) = \lim_{n\to\infty} 1 = 1.$$
