Why must delta be defined in terms of epsilon only?
If you can show that
$$ \forall\epsilon >0 \ \exists\delta>0 \ s.t. 0 < |x-c| < \delta \implies \delta = \frac{\epsilon}{|x+3|}, $$
then you might be able to use $\delta = \frac{\epsilon}{|x+3|}$ in your proof. Can you prove the implication?
The piece above is a broad hint about where you are going wrong. Here is a more explicit explanation:
You can think of the definition of a limit as a game you play against a perfectly clever opponent.
Your opponent moves first by selecting $\epsilon.$ They can choose any positive number whatsoever, tiny, huge, or in between.
Next, it's your turn; you get to select $\delta.$ You can choose any $\delta$ you like as long as it's positive.
Now it's your opponent's move again. They get to choose $x,$ but this time the choice is restricted to a punctured neighborhood of $c$ that you defined when you chose $\delta$; namely, they have to choose $x$ such that $0 < \lvert x - c \rvert < \delta.$
Now if $\lvert f(x) - L \rvert \geq \epsilon,$ you lose.
But to do an actual proof, you need to describe a perfect strategy so that your opponent can never defeat you. In effect, your formula for $\delta$ is a program for a limit-game-playing robot that plays your part of the game. Making your task even more difficult, your opponent will be able to know the program before the game starts. So if there is any flaw in the program, your opponent can exploit it and win. Flaws include values of $\delta$ that let the opponent set $x$ so that $\lvert f(x) - L \rvert \geq \epsilon,$ as well as cases where your rule doesn't produce a definite answer.
So let's say your "program" for the robot is $$ \delta = \frac{\epsilon}{|x+3|}.$$
Your opponent chooses $\epsilon = 10^{-100}.$ Now it's your move. What does your robot choose for $\delta$?
Your rule for $\delta$ requires the robot to know $x,$ which is a number your opponent has not selected yet. So your robot has no way to compute $\delta.$ Poor robot. You lose.
The point is that you are not writting the complete definition of limit. It should be:
$$\forall \varepsilon>0\ \exists\delta>0; \forall x (\textrm{in the domain of}\ f); 0<|x-c|<\delta\Rightarrow |f(x)-L|<\varepsilon.$$
Notice there is a $\forall x$ such that $0<|x-c|<\delta$. This means you must find a delta in such a way that every $x$ which satisfies the condition $0<|x-c|<\delta$ will also satisfy $|f(x)-L|<\varepsilon$.
Geometrically, $0<|x-c|<\delta$ represents an interval centered in $c$ with radius $\delta$. The definition says that for $\varepsilon>0$ fixed there exists such a region which is mapped inside the interval centered in $L$ with radius $\varepsilon$.
Many answers have emphasized what the definition of limit you've given means, i.e. that "$ \forall \epsilon \; \exists \delta$ ..." means that delta must be determinable from epsilon, but it seems none have answered the primary question of why this must be the definition to appropriately capture the notion of a limit. I'll try to address this. First, to established our intuition, the original definition ensures that no matter how small a region around $L$ we choose, we can find some neighborhood of $c$ on which $f(x)$ is always within that small region.
Let's consider what the definition according to your suggested interpretation might look like; that is, for a function $f:U \to \mathbb{R}$ ($U \subset \mathbb{R}$ being the domain of $f$) suppose we define $$\lim_{x \to c} f(x) = L \iff \left( \forall \epsilon>0 \; \exists\,\delta_\epsilon:U \backslash \{c\} \to \mathbb{R}_+ \text{ s.t. } |x-c| < \delta_\epsilon(x) \implies |f(x)-L| < \epsilon \right).$$ There's nothing self-contradictory about this proposed definition at the outset-- one can certainly write this down and choose to work with it. However, it utterly fails to capture what a limit should mean (or much of anything, for that matter), the primary reason being that $|x-c| < \delta_\epsilon(x)$ need not be satisfied by any (let alone all) $x$ sufficiently close to $c$ for arbitrary functions $\delta_\epsilon$. A limit should tell us what $f$ is doing everywhere around $c$, and the condition stipulated by this definition says nothing about those $x$ near $c$ that don't satisfy $|x-c| < \delta_\epsilon(x)$. What's more, this definition makes limits (very) nonunique.
To see these objections more clearly, consider the prototypical step function $$f(x) = \begin{cases} 1, & x \geq 0 \\ 0, & x < 0 \end{cases}$$ I expect we can agree that the limit $\lim_{x \to 0} f(x)$ should not exist, i.e. there is no one number that $f$ tends to "approach" as $x \to 0$. However, given $\epsilon > 0$, consider the function $$\delta_\epsilon(x) = \begin{cases} 2|x|, & x > 0 \\ |x|/2, & x < 0 \end{cases}$$ Then the statement that $|x-0| < \delta_\epsilon(x) \implies |f(x)-1| < \epsilon$ is true, since $|x-0| < \delta_\epsilon(x)$ is only satisfied for $x>0$, where $f$ is identically equal to $1$. That is, according to our proposed definition, $\lim_{x \to 0} f(x) = 1$! A very simple modification also demonstrates that $\lim_{x \to 0} f(x) = 0$. But the situation is even (much) worse than this! By choosing $\delta_\epsilon$ such that $|x-c| < \delta_\epsilon(x)$ is never satisfied (say $\delta_\epsilon(x) = |x-c|/2$), the definition is shown to be vacuously satisfied for any $L$ whatsoever!
We're therefore forced to the following conclusion: allowing $\delta$ to depend on $x$ as well simply breaks the definition's meaning entirely. Placing the necessary restrictions on the functions $\delta_\epsilon$ to make the above definition reasonable will ultimately give you the standard definition.