Here's my question: Is it valid to set ##\delta## to ##\frac{\epsilon}{|y+x|}##? In the case that ##|y+x| \geq 1##, there is no risk of division by zero or assigning ##\delta## to a negative value, so ##\delta## would exist and be positive. If so, then the condition is met.

[Note: This is only the first part of the problem; I do intend to solve the general case with induction.]

I'm afraid it is not valid. You want your ##\delta## to be independent of ##x##. This means that you want to find a ##\delta>0## such that your implication holds for all ##x##.
If you choose ##\delta = \varepsilon / |x+y|##, then you will get a different value of ##\delta## for each ##x##. Clearly this is not what you want.

Is this alright? I notice that there is the possibility that ##\delta = \epsilon< |y-x|## for certain x, y, and ε, which would cause ##|y^2-x^2| < \epsilon## but ##|y-x| \geq \delta##; however, ##|y-x| < \delta \implies |y^2-x^2| < \epsilon## requires only that ##|y^2-x^2| < \epsilon## is true whenever ##|y-x| < \delta## and not the other way around, so it seems okay to me. Am I correct?

For n>2, the following formula is helpful: ##|y^n-x^n| = |y-x||y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}|##

It seems to me that, provided my methods for n=2 were valid, that the same arguments could be used for the cases ##|y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}| < 1## and ##|y^{n-1} + xy^{n-2}+\cdots+x^{n-2}y+x^{n-1}| \geq 1## to show that ##\delta = \epsilon## satisfies all conditions for continuity.