Why shouldn't the definition of a limit say $|f(x) - L| = 0$ instead? The definition of a limit $L$ for function $f$ is as follows. Let  $(a,b)$  be an open interval in $\mathbb{R}$ and $x_0$ such that $x_0 \in (a,b)$. Let $f$ be a real value function defined on all $(a,b)$. It is said that $f$ has a limit $L$ as it approaches point $x_0$ if for every real $\epsilon > 0$, as little as we'd like, there exists a real $\delta$ such that for every $x$ that $0 < \lvert x - x_0 \rvert < \delta$ then $\lvert f(x) - L \rvert < \epsilon$.
But why shouldn't it say instead that there must exist a real $\delta$ such that for every $x$ that $0 < \lvert x - x_0 \rvert < \delta$ then $\lvert f(x) - L \rvert = 0 $?
Let's imagine there doesn't exist such a $\delta$. This essentially means that there is some value (let's call it $\lambda$) such that for every real $\delta_1 > 0$, such that $0 < \lvert x - x_0 \rvert < \delta_1$, $\lvert f(x) - L\rvert \ge \lambda$. If we just choose an $\epsilon$ in which $0 < \epsilon < \lambda$, there won't be a limit to the function as by necessity
$$\lvert f(x) - L \rvert \ge \lambda \implies \lvert f(x) - L \rvert > \epsilon$$
What's wrong with my logic in this? And why do we always bound $\lvert f(x) - L \rvert < \epsilon$ and not equal to zero?
 A: 
But why shouldn't it say instead, that there must exist a real $\delta$ such that (...)

We already have a problem in this simple statement. What do you mean exactly? Because with the previous statement $\delta$ depends on $\epsilon$. But now you got rid of $\epsilon$. So what exactly is your definition? The following?

There exists $\delta > 0$ such that if $0<|x-x_0|<\delta$ then $|f(x)-L|=0$.

So first of all $|f(x)-L|=0$ if and only if $f(x)=L$. On the other hand note that the "$0<|x-x_0|<\delta$" piece is not really related to "$|f(x)-L|=0$" piece. In particular you simply claim that $f(x)=L$, assuming there is $x_0$ with $0<|x-x_0|<\delta$. Note that $f(x)=L$ is true (assuming we define $L:=f(x)$ to begin with) regardless of whether "$0<|x-x_0|<\delta$" condition holds or not. There's just no connection between these two statements. So what exactly does it mean? Well, pretty much nothing. Every function satisfies that, continuous or not. And so the property isn't really useful.
The point of the original definition is that if two arguments are close, then values over them are also close. And so "being close" relationship in the domain and in the range are related and affect each other. And "being close" is formally stated via $|a-b|<c$ inequalities, not $|a-b|=0$ equalities (which simply means $a=b$).
