How do I prove that: $$\lim_{h\to0}f(x_0+h) = \lim_{x\to x_0}f(x)$$ using $\epsilon$-$\delta$ definition of limit.
I have seen a few proofs online for similar questions. But they assume a case such as $x_0+h=x$, or that $\delta$s for both functions are equal. Which I don't understand.
Intuitively they both approach the same point (Assuming the limits exist), so why are such conditions necessary?
One of the proofs I found online: Epsilon-Delta Proof of Limits Being Equal
The method you described is essentially the way. I suspect that your own issues are with using the limit definition itself. As you'll no doubt be aware, to say $\lim_{x \to x_0} f(x) = L$ is to say:
So, no matter what positive number $\varepsilon > 0$ I choose, you should be able to provide me with a $\delta > 0$ so that, whenever $x \neq x_0$ is between $x_0 - \delta$ and $x_0 + \delta$, we have $f(x)$ is between $L - \varepsilon$ and $L + \varepsilon$.
When you're proving a limit, the onus is on you to find such a $\delta$. For most limits, this $\delta$ will change depending on the $\varepsilon$ provided. So, it is natural to have it as a function of $\varepsilon$.
When you're using an established limit, you're allowed to pick whatever positive $\varepsilon$ you want, and can assume that there is a $\delta$ to match (that's what assuming the limit's existence gives you). Then, for any given $x$ satisfying $0 < |x - x_0| < \delta$, you can then assume $|f(x) - L| < \varepsilon$.
So, let's look at the given problem. To start with, we should assume one of the limits exists, and give it a name. We'll stick with $L$, for simplicity. Let's assume that $\lim_{h \to 0} f(x_0 + h) = L$, and try to prove that $\lim_{x \to x_0} f(x) = L$ as well.
To prove the latter limit, we have suppose we have some arbitrary $\varepsilon > 0$, and find a way to construct a $\delta > 0$ such that, whenever $0 < |x - x_0| < \delta$, we also have $|f(x) - L| < \varepsilon$. So, the onus is on us to provide the $\delta$.
We are assuming that $\lim_{h \to 0} f(x_0 + h) = L$. That is, if we set $\varepsilon'$ to any strictly positive number we wish, we can automatically assume that there is some $\delta' > 0$ such that, whenever $0 < |h - 0| < \delta'$, we also have $|f(x_0 + h) - L| < \varepsilon'$.
How do we do it? We must, as usual, start by supposing $\varepsilon$ is some arbitrary strictly positive number. We have no control over its value, but we know it's positive.
Next, we use the definition of $\lim_{h \to 0} f(x_0 + h) = L$, taking $\varepsilon' = \varepsilon$ (we can set $\varepsilon'$ to any positive number, so why not $\varepsilon$?). We then get a $\delta'$ such that $$0 < |h - 0| < \delta' \implies |f(x_0 + h) - L| < \varepsilon' = \varepsilon. \tag{$\star$}$$ Remember, we need to find a $\delta$ such that $$0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon.$$ I claim (as do all the other answers you've seen) that $\delta = \delta'$ (the value with the certain properties that we are allowed to assume exists) will do the trick.
How do we prove this claim? Start by assuming $0 < |x - x_0| < \delta = \delta'$. Recall that $(\star)$ tells us that, whenever any real number $h$ satisfies $0 < |h| < \delta'$, then we automatically have $|f(x_0 + h) - L| < \varepsilon'$. So, our assumption says that the particular value of $h$, $x - x_0$, satisfies this inequality. Thus, using $(\star)$ in the particular case $h = x - x_0$, we get $$0 < |x - x_0| < \delta = \delta' \implies |f(x_0 + (x - x_0)) - L| < \varepsilon' = \varepsilon.$$ Once we simplify the right hand side, this is precisely the implication we wanted to prove. Thus, choosing $\delta = \delta'$ works; we have a guaranteed $\delta > 0$ such that $$0 < |x - x_0| < \delta \implies |f(x) - L| < \varepsilon.$$ Technically speaking, we aren't done yet, as we should also assume that $\lim_{x \to x_0} f(x)$ exists, and use that to prove $\lim_{h \to 0} f(x_0 + h)$ exists and has the same value, but I'll leave that to you.