I'm afraid that your negation of continuity is incorrect. First, let us remember the definition of "continuous at $a$":
$\forall \epsilon\gt 0\Biggl( \exists \delta\gt 0\biggl(\forall x\bigl( |x-a|\lt \delta \Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\biggr)\Biggr)$.
Now, how do we write down the negation? (I'm assuming that "discontinuous at $a$" means "defined at $a$ but not continuous, by the way)?
Let's break it down: what is the negation of a "For all" statement?
The negation of $\forall x (P(x))$ is $\exists x(\neg P(x))$.
That is, the negation of "For every $x$, $P(x)$ is true" is "There is at least one $x$ for which $P(x)$ is false". So, the negation of the definition of continuity will be:
$\exists \epsilon\gt 0\Biggl(\neg\Biggl(\forall \delta\gt 0 \biggl(\forall x\bigl( |x-a|\lt \delta \Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\biggr)\Biggr)\Biggr)$.
Now we need to find the negation of the statement inside the "For all", which is a statement of the form "There exists..." What is the negation of a "There exists" statement?
The negation of $\exists y(Q(y))$ is $\forall y(\neg Q(y))$.
That is: the negation of "There exists a $y$ such that $Q(y)$ is true" is "For every $y$, $Q(y)$ is false". So, to negate that "There exists" statement, we have:
$\exists \epsilon\gt 0\forall\delta\gt 0\Biggl(\neg\Biggl(\forall x\bigl( |x-a|\lt \delta\Rightarrow |f(x)-f(a)|\lt \epsilon\bigr)\Biggr)\Biggr)$.
Now we need to negate a "for all" statement. We already know how to do that:
$\exists \epsilon\gt 0 \forall\delta\gt 0 \Biggl( \exists x\Bigl( \neg\bigl( |x-a|\lt\delta\Rightarrow |f(x)-f(a)\lt\epsilon\bigr)\Bigr)\Biggr)$.
Now we need to find the negation of the implication. What is the negation of an implication?
The negation of $A\Rightarrow B$ is $A\text{ and }\neg B$.
That is, the negation of "If $A$, then $B$" is "$A$, and not $B$." So:
$\exists \epsilon\gt 0 \forall \delta\gt 0\Biggl( \exists x\bigl( |x-a|\lt\delta \text{ and }\neg(|f(x)-f(a)|\lt\epsilon)\bigr)\Biggr)$.
Finally, we need to find the negation of "$|f(x)-f(a)|\lt\epsilon$". That's $|f(x)-f(a)|\geq \epsilon$. So we finally get:
$\exists \epsilon\gt 0\forall\delta \gt 0 \Biggl( \exists x\bigl( |x-a|\lt\delta\text{ and }|f(x)-f(a)|\geq \epsilon\bigr)\Biggr)$.
In words: "there is an $\epsilon\gt 0$ such that, no matter what $\delta\gt 0$ you pick, there is an $x$ which is $\delta$-close to $a$, but with $f(x)$ not $\epsilon$-close to $f(a)$."
Note. Some people say that $f$ is discontinuous at $a$ if and only if it is defined at $a$ but not continuous at $a$. Less common is to say that $f$ is discontinuous at $a$ if and only if it is not continuous at $a$. Under the latter (in my experience uncommon) definition, "$f$ is discontinuous at $a$" would be a disjunction between the formula above, and "$f$ is not defined at $a$".