Lemma: Given any function $f:M\to N$ where $M$ and $N$ are both metric spaces, $\lim_{x\to a}f(x)$ converges to $L$ only if given any function $\gamma:\mathbb{I}\to M$ (where $\mathbb{I}$ is the unit interval) such that $\lim_{t\to1}\gamma(t)=a$, $\lim_{t\to1}f(\gamma(t))=L$.
I've seen the above lemma used implicitly in various analysis proofs without acknowledgement several times. My attempt to prove it goes as follows:
Assume that the lemma was false. Then all possible $\lim_{t\to1}f(\gamma(t))$ converge to $L$, but $\lim_{x\to a}f(x)$ does not. Then, by the definition of limits, there is an $\epsilon>0$ such that for all $\delta$ there exists an $x$ such that $d(x, a)<\delta$ and $d(f(x), L)\ge\epsilon$. Denote the set of all such $x$ as $C_\delta$. Define $\gamma:\mathbb{I}\to M$, $\gamma(x)=$ some element in $C_{1-x}$. By the axiom of choice such a function exists. Obviously $\lim_{t\to1}\gamma(t)=a$, but $\lim_{t\to1}f(\gamma(t))\not=L$. This contradictions our assumptions. Thus, by contradiction, $\lim_{x\to a}f(x)=L. \square$
This proof uses the axiom of choice, which is quite unsatisfactory. Does anyone know if this theorem can be proven without using the axiom of choice? I'm fine with assuming that M and N are complete, although not much more than that.