Sometimes people say that the concept of a continuous map express, intuitively, that "small changes in the domain must yield small changes in the respective images, in the codomain".
However, this is not the definition we actually use. We say instead: "If one wants small changes in the image, say, by $\epsilon$, one can find a $\delta$ (depending on $\epsilon$), such that, changes in the domain of at most $\delta$, reflects small changes in the images, by the $\epsilon$ previously given."
So, thinking of the intuition given above, consider the following definition.
A function $f:\mathbb{R}\to \mathbb{R}$ is said to be continuous at $x\in \mathbb{R}$ if for all $\delta>0$, there is a $\epsilon(\delta)$ such that $f((x-\delta,x+\delta))\subset (f(x)-\epsilon,f(x)+\epsilon)$.
Ok, so here I'm just switching the symbols of $\forall$ and $\exists$ in the usual definition of continuity. This definition is clearly bad, because, given a $\delta$, I can get a tremendously big $\epsilon$ and many functions that we don't want to be continuous would be it, with this definition.
But, what if we add to that definition:
A function $f:\mathbb{R}\to \mathbb{R}$ is said to be continuous at $x\in \mathbb{R}$ if there is a (choice) function $\epsilon_x:\mathbb{R}_+\to \mathbb{R}_+$ such that:
(1) $f((x-\delta,x+\delta))\subset (f(x)-\epsilon_x(\delta),f(x)+\epsilon_x(\delta))$, for all $\delta>0$;
(2) $\lim\limits_{\delta\to 0^+}\epsilon_x(\delta)=0$.
We say that $f$ is continuous if $\epsilon_x$ exists for all $x\in \mathbb{R}$.
Now, I'm saying that, when $\delta$ goes small, so does $\epsilon$. Is this definition still bad?
The next step would be to check if this definition implies the usual one. Well, let $\epsilon>0$ be given. By (2), there is a $\delta'>0$ such that, $0<\delta<\delta'\implies 0<\epsilon_x(\delta)<\epsilon$, and using $\delta'$ in the usual definition allows us to conclude that $f$ is continuous at $x$ in the usual sense.
So, are these definitions the same? I'm a little confused about the converse... (it smells like axiom of choice...)
As you showed, if $f$ satisfies your definition it is continuous at $x$. Conversely, suppose $f$ is continuous at $x$. We may take $$\epsilon_x(\delta) = \sup_{t \in [x-\delta, x+\delta]} |f(x)-f(t)| $$ (using the fact that a continuous function on a closed interval has a maximum there) and your condition will be satisfied. No axiom of choice needed here.