$$
|x- x_0| \lt 1 \\
|x| \lt 1+ |x_0|
$$
Now, let's manipulate this thing
$$
|xy - x_0 y_0| = \big| |x| |y-y_0| + |y_0| |x-x_0| \big| \lt |x| |y-y_0| + |y_0||x-x_0|\\
\text{If we want} \\
|x| |y-y_0| + |y_0| |x-x_0| \lt \epsilon \\
\text{then we need} \\
|x| |y-y_0| \lt \frac{\epsilon}{2} \implies |y-y_0| \lt \frac{\epsilon}{ 2( 1+x_0)} \\
|y_0| |x- x_0| \lt \frac{\epsilon}{2} \implies |x-x_0| \lt \frac{\epsilon}{2 |y_0|}
$$

@CalvinKhor (When you come back) So, I have two questions:
1. The actual answers are
$$
|x-x_0| \lt ~min~\left( 1, \frac{\epsilon}{2 (|y_0 +1 )}\right) \\
|y-y_0| \lt \frac{\epsilon}{2 (|x_0| +1)} $$
Why my $|x-x_0|$ doesn't match with the answer?

2. Why we assumed $|x-x_0| \lt 1$ ?

1. The reason is because there are an uncountable number of different proofs of the same result, just like earlier, each $k$ gave a different proof even though you chose $k=1/2$
2. Did you use it? Then that's why. I don't understand the question. You could also use a different but similar identity $xy - x_0 y_0 = y(x-x_0) + (y-y_0)x_0$ and now you won't assume $|x-x_0|<1$ but you'll need to assume something else. Again, over a million different proofs.