No. The point is that some continuous functions do not have this property! So the two statements are not equivalent.
Consider any constant real function $f$: let $x,y$ be any two points and $\epsilon >0$. Then $|f(x)-f(y)|=0<\epsilon$, but, we can make $|x-y|$ "as large as we want". For instance, take $x=0$ and let $y$ go to infinity: no $\delta$, in this case, works.
There is no "sure fire" way of proving continuity of a function. However, the steps are usually a bit backward to what the actual definition is. That is, the definition says that $f$ is continuous at $a$ if for each $\epsilon>0$, there exists $\delta >0$ such that if $|x-a|<\delta$, then $|f(x)-f(a)|<\epsilon$.
We start the proof by taking an arbitrary $\epsilon > 0$. However, we then usually do not magically think of a $\delta$ that would fit. What we typically try to do is simplify the expression $|f(x)-f(a)|$ and prove that it is "small", keeping in the back of our mind that we can always make $|x-a|$ "small".
In your case, for example, we can prove first that $|f(x)-f(a)|=|\frac{x-a}{\sqrt x + \sqrt a}|$. Now, we want to ask ourselves: If $|x-a|$ is small, is this expression also small?
This is done by trying to find a small upper bound for the expression that will hold whenever $|x-a|<\delta$. In your particular case, this is fairly simple, since we know that the expression is smaller than $\frac{|x-a|}{\sqrt a}$, and this is smaller than $\epsilon$ if $\delta$ is set small enough.
Now, once we did these steps, we take a step back, and think about what we just did.
Looks like we found our $\delta$, and we have to set our $\delta=\epsilon\sqrt a$.
Hmm, is this OK? Well, as long as $a>0$, we know that $\delta>0$, and we have proven that if $|x-a|<\delta$, then $|f(x)-f(a)|<\frac{\delta}{\sqrt a}$, ,which is equal to $\epsilon$, so it looks like we are almost done.
Almost, because we can see that if $\sqrt a=0$, then our proof does not work! We need to do that part separately. If $a=0$, then $|f(x)-f(a)|=|\sqrt x|$. We still need to prove that if $|x-0|$ is small, then $|\sqrt x|$ is also small.
Best Answer
We show that the given function is continuous at any point $(a,b)$. So given any $\epsilon\gt 0$, we want to produce a number $\delta$ such that $$|(x+y)^2-(a+b)^2|\lt \epsilon$$ whenever $d((x,y),(a,b))\lt \delta$. Here $d((x,y),(a,b))$ is the Euclidean distance between $(x,y)$ and $(a,b)$, that is, $\sqrt{(x-a)^2+(y-b)^2}$. Note that if $\sqrt{(x-a)^2+(y-b)^2}\lt \delta$ then $|x-a|\lt \delta$ and $|y-b|\lt \delta$.
We have $$(x+y)^2-(a+b)^2=((x+y)-(a+b))(x+y+a+b).\tag{1}$$ By choosing $\delta$ to be small, we have very direct control over the size of $|(x+y)-(a+b)|$, and can force it to be as small as we wish. But we must also get control over the size of $|x+y+a+b|$, to make sure it does not get too big. That is what the rest of the calculation is devoted to.
Suppose that $|x-a|\lt \delta$ and $|y-b|\lt \delta$ and $\delta\lt 1$. Then $|x|\lt |a|+1$, and $|y|\lt |b|+1$. It follows that $|x+y+a+b|\lt 2+2|a|+2|b|$. Thus from (1) we have $$|(x+y)^2-(a+b)^2|\lt (2\delta)(2+2|a|+2|b|).\tag{2}$$ It follows from (2) that if $0\lt \delta\lt \frac{\epsilon}{2(2+2|a|+2|b|)}$ then $|(x+y)^2-(a+b)^2|\lt \epsilon$.