Prove $\lim_{x\to 0} f(x)$ exists $\implies\lim_{x\to 0} f(x^2)$ exists

(this is the part I was most questionable about - am I allowed to do this?).

You are. But if it makes you uncomfortable skip it and work back.

$\lim_{x\to 0} f(x) = L$ is given. so: For any $\epsilon>0$ there is a $\delta>0$ so that $0<|x| < \delta \implies |f(x) - L| < \epsilon$.

And $\lim_{x\to 0}x^2 = 0$ because for any $\epsilon >0$ there is a $\delta=\sqrt \epsilon> 0$ so that $0<|x|<\delta\implies |x^2| < \delta^2 =\epsilon$.

So for any $\epsilon$ let $\delta_{inbetween}$ be a value so that $0< |x|<\delta_{inbetween} \implies |f(x)-L| < \epsilon$ and let $\delta$ be a value so that $0<|x|<\delta \implies |x^2| < \delta_{inbetween}$.

Then $0< |x|< \delta \implies 0< |x^2| < \delta_{inbetween} \implies |f(x^2)-L| < \epsilon$.

So $\lim_{x\to 0} f(x^2) = L$.