I'm refreshing my topology knowledge from the first chapter of Manifolds, Tensor Analysis, and Applications by Marsden, Ratiu, and Abraham. I attempted a proof at the following question, but I'm not totally sold on whether it works:
Let $X,Y$ be topological spaces and $\sim,\approx$ be equivalence relations on $X$ and $Y$, resp. Let $\phi: X \to Y$ be continuous such that $x_1 \sim x_2$ implies $\phi(x_1) \approx \phi(x_2)$. Show that the induced mapping $\hat{\phi}: X/\sim \to Y/\approx$ is continuous.
Here is my attempt at a proof:
Let $U \subset Y/\approx$ be an open set, and let $\pi_\sim: X \to X/\sim,\, \pi_\approx : Y \to Y/\approx$ be the canonical projections. First, note that $\pi_\approx^{-1}(U)$ is open in $Y$ by the quotient topology on $X/\sim$. Since $\phi:X \to Y$ is continuous, then $(\pi_\approx \circ \phi)^{-1}(U)$ is open in $X$; that is, $\pi_\approx \circ \phi : X \to Y/\approx$ is continuous. Since $\pi_\approx\circ \phi = \hat{\phi}\circ\pi_\sim$, $$ (\pi_\approx \circ \phi)^{-1}(U) = (\hat{\phi}\circ\pi_\sim)^{-1}(U) = \pi_\sim^{-1}\circ\hat{\phi}^{-1}(U) $$ is open, and by the definition of the quotient topology on $X/\sim$, $\hat{\phi}^{-1}(U)$ is open. Therefore, $\hat{\phi}$ is continuous.
This seems to work, but I'm not sure what to do with the hypothesis "$x_1 \sim x_2 \implies \phi(x_1) \approx \phi(x_2)$". Is this just to define $\hat{\phi}$ via $\pi_\approx\circ \phi = \hat{\phi}\circ\pi_\sim$, or am I missing something?
You do not miss anything. As you said, the condition $$(*) \quad x_1 \sim x_2 \implies \phi(x_1) \approx \phi(x_2)$$ implies that $\hat \phi([x]_\sim ) = [\phi(x)]_\approx$ is well-defined.
You can reformulate this as follows: Condition $(*)$ implies that there exists a function $\hat \phi : X/\sim \phantom{} \to Y/\approx$ such that $\pi_\approx\circ \phi = \hat{\phi}\circ\pi_\sim$ (actually $(*)$ is equivalent to the existence of $\hat \phi$). Note that $\hat\phi$ is uniquely determined because $\pi_\sim$ is surjective.
Concerning the continuity of $\hat \phi$ you may proceed a little bit shorter. Since $\phi$ and $\pi_\approx$ are continuous, also their composition $\pi_\approx\circ \phi$ is continuous. Your next argument correctly establishes the continuity of $\hat \phi$. However, here you just reprove the well-known universal property of quotient maps: If $p : X \to X'$ is a quotient map (which means that $p$ is surjective and $U \subset X'$ is open iff $p^{-1}(U) \subset X$ is open), then the following holds true:
Given any continuous function $f : X \to Z$ and any function $f' : X' \to Z$ such that $f' \circ p = f$, then $f'$ is continuous.