Continuity of $f(x)=x^p$ when $p$ is a real number and $x\in (0,\infty)$ Here is my final answer. 
Definition Let $x>0$ be a real, and $\alpha$ be a real number. We define the quantity $x^{\alpha}$, by the formula $\text{lim}_{n\rightarrow\infty} x^{q_n}$ where $(q_n)$ is a sequence of rationals which converges to $\alpha$. (I've shown that the definition is well-defined).
Proposition: Let $p$ be a real number. Then the function $f: (0,\infty)\rightarrow \mathbb{R}$ defined by $f(x)=x^p$ is continuous. 
Proof: Let $x_0\in (0,\infty)$ we have to show  $\lim_{x\rightarrow x_0, x\in(0,\infty)} f(x)=f(x_0)$.
(1) Suppose $x_0=1$. 

Claim 1: For all natural numbers $\lim_{x\rightarrow 1, x\in\mathbb{R}} x^n=1$. Let $n=0$, so $x^n=1$ which is trivial. Suppose we have  proven the assertion for $n\ge 0$. So, $x^{n+1}=x^nx$ and then 
$$\lim_{x\rightarrow 1, x\in\mathbb{R}} x^{n}x=\lim_{x\rightarrow 1, x\in\mathbb{R}} x^{n}\cdot\lim_{x\rightarrow 1, x\in\mathbb{R}} x=1$$

2) Now we have to show that $\lim _{x\rightarrow 1; x\in (0, \infty)} x^p = 1$.
Let  $(x_n)_{n=0}^\infty$ be a sequence of positive real numbers which converges to $1$. We' like to show that $(x_n^p)\rightarrow 1$. 
Let $\varepsilon>0$ be arbitrary and let choose some $m\in \mathbb{N}$ so that $m> p$. Since $(1+1/k)_{k=1}^\infty$ and $(1-1/k)_{k=1}^\infty$ both converges to $1$, using the claim $1$, $(1+1/k)^m$ and $(1-1/k)^m$ converge also to $1$. Let $K_\varepsilon$ be a natural number such that both sequences are $\varepsilon$- close to $1$ for any $k\ge K_\varepsilon$. Let us fix some $k$, so that  $k\ge K_\varepsilon$ and $1-1/k > 0$.
Since $(x_n)$ converges to $1$, there is some $N_{1/k}$ such that $|x_n-1|\le 1/k$ for all $n\ge N_{1/k}$, i.e., $1-1/k\le x_n\le 1+1/k$. So, $(1-1/k)^p\le x_n^p\le (1+1/k)^p$. Also we know that $1+1/k>1$ and $p<m$, thus $(1+1/k)^p<(1+1/k)^m$. Similarly $1-1/k<1$, thus  $(1-1/k)^p>(1-1/k)^m$. Putting all the inequalities together we have 
$$(1-1/k)^m< x_n^p< (1+1/k)^m $$
Since both $(1+1/k)^m $ and $(1-1/k)^m $ are $\varepsilon$-close to $1$, hence $x_n^p$ is. Thus $(x_n)^p$ converges to $1$ as desired. Now since $(x_n)$ was an arbitrary sequence of positive real numbers converging to $1$, hence the result  hold for any sequences. Therefore  $\lim _{x\rightarrow 1; x\in (0, \infty)} x^p = 1$.
3) Let $x_0 \in (0 \infty)\backslash\{1\}$ and let $(x_n)_{n=0}^\infty$ be a sequence of positive real numbers which converges to $x_0$. Using the limit laws we know that $x_n/x_0$ converges to $1$ and so by part 2), we have $(x_n/x_0)^p \rightarrow 1$. Thus
\begin{align}\lim_{n\rightarrow \infty}x_n^p=\lim_{n\rightarrow \infty} x_0^p (x_n/x_0)^p\\
= x_0^p \lim_{n\rightarrow \infty} (x_n/x_0)^p \end{align}
and since $\lim_{n\rightarrow \infty} (x_n/x_0)^p=1 $, hence $\lim_{n\rightarrow \infty}x_n^p=x_0^p$. Since $(x_n)$ was an arbitrary sequence of positive real numbers converging to $x_n$, this would imply that $f$ is continuous on $(0,\infty)$ as desired.
Thanks to anyone. 

The book's hint is spot on: suppose you have proven that $x^p$ is continuous at $1$; and choose some $\alpha>0$ distinct from $1$. Prove that $x^p$ is continuous at $\alpha$ iff $\alpha^{-p}x^p$ is continuous at $\alpha$, and note that this last is equivalent to $$\left(\frac x\alpha\right)^p\to 1$$ as $x\to \alpha$. But $x/\alpha\to 1$ as $x\to \alpha$, and since $t^p$ is continuous at $t=1$; we're done! You do have to prove $(xy)^p=x^py^p$ for $x,y$ real numbers first. 
