You are right that $\frac xx$, or rather the function conventionally meant by that term, is continuous. But I think there is a conceptual issue at play here, and a common one. Namely, that the functional equation
$$f(x)=\textrm{some expression in }x$$
already determines the function. This is not true. When building a function, we specify things in this order: domain and codomain first, functional equation second. That is, we choose the domain and codomain of the function, and then we are bound to give a functional equation which is actually defined for all elements of the domain, and returns only elements of the codomain. In mathematical notation you'll see it like $f:X\to Y,~f(x)=\dots$, where $X,Y$ are domain and codomain, and they are specified first. Now this said, $f:\mathbb R\to\mathbb R,~f(x)=\frac xx$ is not a well-defined function, since the functional equation isn't meaningful for all elements of the domain. However, $f:\mathbb R\backslash\{0\}\to\mathbb R,~f(x)=\frac xx$ is well-defined, since the functional equation is meaningful for all elements of the previously specified domain, and it returns only elements of the previously specified codomain. So to be sensible, when we write "the function $\frac xx$", we usually mean that as shorthand for the latter definition of a function. But technically speaking, just the term $\frac xx$ does not define a function, since we haven't specified domain or codomain, which we must, unless we want our readers to guess (which isn't always unfeasible, to be fair).
Now with all this about the definition of functions out of the way, continuity can be treated more clearly: a function $f:X\to Y$ is continuous at $x\in X$ if [insert your favorite definition of continuity at a point]. A function $f:X\to Y$ is continuous if it is continuous at all $x\in X$.
Notice that this definition only talks about points inside the domain $X$, at which the function must be defined in order to be a function. Whatever lies outside of the domain is of no concern for the definition of continuity.
As a closing remark, the often cited example of the discontinuous function $\frac 1x$ is just bad, because $f:\mathbb R\to\mathbb R,~f(x)=\frac 1x$ is not a function, while $f:\mathbb R\backslash\{0\}\to\mathbb R,~f(x)=\frac 1x$ is a perfectly continuous function. There just isn't any sensible (meaning non-contrieved) way to interpret $\frac1x$ as something that is both a function and discontinuous.
Your proof is the same as that one and is correct, rigorous, and well-written. See also this post and many others, linked or related.
Your note after the definition of $\alpha$-continuity and your remark after lemma 1 are not convincing because if $\forall y,z\in V_\delta(x)\quad|f(y)-f(z)|<\alpha,$ then in particular $\forall y\in V_\delta(x)\quad|f(y)-f(x)|<\alpha,$ and conversely, if $\forall y\in V_\delta(x)\quad|f(y)-f(x)|<\alpha,$ then $\forall y,z\in V_\delta(x)\quad|f(y)-f(z)|<2\alpha$ (this is the proof your lemma 2). To me, the true "difference from standard continuity" is that it is an intermediate notion with $\alpha$ fixed, so that (for $f$ at $x$) standard continuity is equivalent to $\alpha$-continuity for all $\alpha>0$ (this is the statement of your lemma 2).
For me, the shorter the better but it is a matter of taste, so I will not propose an alternative shortened version of yours. Two minor improvements could be to replace "$\alpha$-continuous for $\alpha=1/n$" by "$1/n$-continuous" (and similarly with "$\alpha$-continuous for $\alpha=k$"), and to choose earlier a notation for the set of points where $f$ is (or is not) $\alpha$-continuous, so as to use it in your lemmas and main proof.
The converse is true in $\Bbb R$, i.e. any $F_\sigma$ subset of $\Bbb R$ is the set of discontinuities of some function $f:\Bbb R\to\Bbb R.$
Your theme is related to Oscillation and Borel hierarchy.
Edit: An important point which I did not notice before: The domain of your $f$ should not be $A\subset\Bbb R,$ but $\Bbb R$ itself. Or if you insist on keeping $f:A\to\Bbb R,$ the result will be an $F_\sigma$-subset not of $\Bbb R$ but of $A$, equipped with the induced topology.
Best Answer
A function can only be continuous or discontinuous at points in its domain. So I think the comment responding to you is correct. Think of it in this way: Suppose you have a function $f:\mathbb{R}\to\mathbb{R}$. It wouldn't make sense to say that $f$ is discontinuous at $i\in\mathbb{C}$, even though $\mathbb{R}\subset\mathbb{C}$. In the same way, if $f$ is a function on, say $[0, 1]$, then you can't say that $f$ is discontinuous at $x=10$ for example.
Having said that, I think this borders on a philosophical difference. The difference between "$f$ is neither discontinuous or continuous at $a$ if $f$ is not defined at $a$" and "$f$ is discontinuous at $a$ if it's not continuous at $a$" I think is unlikely to cause a major divergence in understanding for students.