I realize this is an old question, but I stumbled upon it and I felt that the answers don't address an important point in your question. Namely is your reasoning sound? Given that this was three years ago, you may be well aware of what I am going to say, but for the sake of anyone else reading the question and answers, I'll add this.
The answer to this is no, though perhaps it's a nitpick, but let me explain.
Your problem with the, let's call it $\delta$-$\epsilon$ definition you've come up with is that it seemed to imply that the function must be defined on all of $\Bbb{R}$. However there is a subtlety here, we must be careful about the quantification of $x$. While in high school calculus, we say things like the square root function is from $\newcommand{\RR}{\Bbb{R}}\RR\to\RR$, but its undefined when $x<0$. However in mathematics, functions are always defined on their entire domain. Thus we say that $f(x)=\sqrt{x}$ is a function $f:\RR_{\ge 0}\to\RR$. I.e. the square root function is a function defined only on positive reals. This is important because in the usual definition of continuity, $x$ is quantified over the domain of the function. I.e. we say a function from a subset $D\subseteq\RR$ to $\RR$ is continuous at $x_0\in D$ if
$$\forall_{\epsilon >0}\exists_{\delta >0}\forall_{x\in D}|x_0-x|<\delta \implies |f(x_0)-f(x)|<\epsilon.$$
Thus in order for your definition to match the usual notion of continuity, it would have to read
$$\forall_{\delta >0}\exists_{\epsilon >0}\forall_{x\in D}|x_0-x|<\delta \implies |f(x_0)-f(x)|<\epsilon.$$
Then because $x$ is required to be in the domain of the function by definition, the fact that we can choose $\delta$ as big as we want doesn't force any element of $\RR$ to be part of the domain of $f$.
As for why it's important that we restrict $x$ to be in the domain of $f$ for the usual definition of continuity, well for one thing, we'd like to say that the square root function is continuous at 0, which would be impossible if we allowed $x$ to be any real number less than a distance of $\delta$ from $x_0$, since for any $\delta$, $-\delta/2$ is a real number less than a distance of $\delta$ from 0, but $-\delta/2$ is negative, and hence not in the domain of $\sqrt{x}$.
Finally, just some comments on the $\delta$-$\epsilon$ definition you've described (modified so that $x$ has to be in the domain of the function). Let's say a function satisfying the property you've described is called bounded on finite intervals around $x_0$
Some properties:
1) If $f$ is bounded on finite intervals around $x_0$ for some particular $x_0$, it is bounded on finite intervals around $x$ for any $x$ in the domain, $D$, at all.
Proof: Suppose we are given $\delta$, let $\delta'=\delta+|x-x_0|$. Then since $f$ is bounded on finite intervals around $x_0$ it is bounded on $(x_0-\delta',x_0+\delta')\supseteq (x-\delta,x+\delta)$. Hence $f$ is bounded on finite intervals around $x$.
2) $f$ is bounded on finite intervals around $x_0$ for some particular $x_0$ if and only if $f$ is bounded on all bounded subsets of its domain. (Proof is similar to the last property)
3) If $f$ is continuous with closed domain then $f$ is bounded on all bounded subsets of its domain. Thus it is bounded on finite intervals around $x_0$ for any $x_0$ in its domain by 2 and 1.
Proof. Let $B$ be a bounded subset of $D$. Thus $\bar{B}\subset D$, and $\bar{B}$ is compact. Since $f$ is continuous it is bounded on compact sets, and hence on $\bar{B}$. Thus $f$ is bounded on $B$.
4) Not all continuous functions are bounded on bounded subsets of their domain though. As an example $\frac{1}{x}$ is unbounded on $(0,1)$.
