Remember the definition of limit,
$$\forall \epsilon\; \exists \delta : |x-a|<\delta \rightarrow |f(x) - L|<\epsilon $$
Now if $L$ is not the limit negate the definition, that is
$$ \exists \epsilon\; \forall \delta: \;\; \exists x \text{ s.t. } |x-a| <\delta\text{ and } |f(x) - L| \geq \epsilon $$
In your case, as some people already mentioned take $\epsilon =1/2$, now for ever $\delta$ we need to find an $x$ such that $|x-4|<\delta$ and $|9-x-4|\geq 1/2$. Finding such $x$ isn't that difficult, just take $0<r<\min\{\delta, 1/2\} $ and consider $x=4+r$. To check we substitute in the inequalities:
$$|x-4|=|r|<\delta$$
And
$$|9-x-4|=|1-r|>1/2= \epsilon $$
(sorry for the awful format, I'm on my phone)
Just to show that it is not at all hard or confusing to show that $f \colon [0..1) → ℝ,~x ↦ \frac 1 {1 - x}$ is not uniformly continuous using the $ε$-$δ$-definition:
Since $(0..1] → [0..1),~x ↦ 1 - x$ is uniformly continuous and uniform continuity is stable under composition, it suffices to show that $f \colon (0..1] → ℝ,~x ↦ \frac 1 x$ is not uniformly continuous.
For this, fix $ε = 1$. Then for any $δ > 0$ with $δ < 1$, let $x = δ$ and $x' = \frac δ 2$. Then clearly $|x - x'| < δ$, whereas $|1/x - 1/x'| = 1/δ ≥ 1 = ε$. For $δ ≥ 1$, choose $x = 1$ and $x' = \frac 1 2$.
Now regarding your question. To show the equivalence of the $ε$-$δ$-definition and the sequential definition of continuity for metric spaces, one needs the Axiom of Countable Choice. (And to show the equivalence of the generalized neighbourhood definition and the net definition, one needs the full axiom of choice, I think.)
That is to say, the equivalence is non-constructive. This already makes it clear that neither approach is in itself more useful for all cases, but it’s the equivalence of these approaches that is really useful: Thinking non-constructively is hard and having two languages to talk about the same phenomena from two constructively separated places that have already been linked by a bridge saves one the effort of building a new bridge every time it seems handy to switch perspectives.
Philosophically, I’d say that the $ε$-$δ$-perspective of continuity and uniform continuity capture these notions more directly, whereas the perspective on these from sequences and net is more indirect, as if to say: “There is no counterexample of sequences such that they are converging here, but not there. Such a wild thing does not exist!” This makes this perspective perfect for the realm of counterexamples. Proofs using sequences tend to be indirect proofs, using contradiction. It’s often about supposing something isn’t true and then, often non-constructively, choosing a counterexample sequence.
The $ε$-$δ$-perspective, however, not only generalises more easily to more abstract spaces, I also find it to be easier to think about conceptually. For example, I did have an absurdly hard time thinking about how to prove that continuous maps are uniformly continuous on compact sets when thinking about this sequentially, whereas it’s crystal clear from the $ε$-$δ$-perspective. Maybe it’s a matter of socialisation, but it’s easier for me to see things from this perspective. Another example would be that open subgroups of compact topological groups (which are uniform spaces) are exactly closed subgroups of finite index. It’s obvious from this perspective, not so much from the other.
In short: Topological spaces can be really weird, making set-theoretic topology a study of weird things. If you want to do neat topology, then it might become a study about how some weird things are not happening. That makes the sequential approach a bit more powerful, albeit less conceptual.
And whenever you encounter an extension of this equivalence such as the equivalence of compactness and sequential compactness, expect proofs to be non-constructive.
Best Answer
I'm assuming your true concern is understanding $\epsilon$,$\delta$-proofs. We first go over the definition of continuity at a point, then give an example.
The Definition
Let $f$ be a function from some subset of $\mathbb{R}$, call it $D$ for domain, to another subset of $\mathbb{R}$, call it $C$ for codomain. That is,$$f:D\subseteq\mathbb{R}\to C\subseteq\mathbb{R}$$ Let $a$ be an arbitrary element of the domain. That is, $$a\in D$$ $f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in D$, if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
An Example
Let $f:[3,\infty)\to \mathbb{R}$ be defined by $$f(x)=\sqrt{2x-6}$$ for every $x\in[3,\infty)$. (It is important to define the function explicitly; that is, the domain is relevant for this proof.)
$f$ is continuous at $4$.
Proof. To prove $f$ is continuous at $4$, we will show that for each $\epsilon>0$, there exists $\delta>0$ such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$.
Let $\epsilon>0$ be arbitrary. Notice for each $x\in[3,\infty)$, $$|f(x)-f(4)|=\sqrt{2}|\sqrt{x-3}-1|=\sqrt{2}\left|\frac{x-4}{\left(\sqrt{x-3}\right)+1}\right|<\sqrt{2}|x-4|\tag{1}$$ Define $\delta\overset{\text{def}}{=}\dfrac{\epsilon}{\sqrt{2}}$. It is important to understand why we defined $\delta$ this way (in relation to $\epsilon$). Anyway, we continue with the proof. For each $x\in[3,\infty)$, if $|x-4|<\delta$, then $$|f(x)-f(4)|\overset{(1)}{<}\sqrt{2}|x-4|<\sqrt{2}\cdot\delta=\epsilon$$ Recall $\epsilon>0$ was arbitrary. Therefore for each $\epsilon>0$, there exists $\delta>0$ (namely $\epsilon/\sqrt{2}$), such that for each $x\in[3,\infty)$, if $|x-4|<\delta$, then $|f(x)-f(4)|<\epsilon$. Therefore $f$ is continuous at $4$.$\square$