Suppose that $X$ is a locally connected and simply connected space and $f:X\to Y$ is a function such that for every path $\phi:[a,b]\to X$ the composition $f\circ\phi$ is continuous. Does it follow that $f$ is continuous? Here is a proof using countable choice if we additionally assume that $X$ is first-countable and locally path connected:
Assume that $f$ is not continuous at $x\in X$, and let $(U_n)$ be a local base at $x$ which forms an inclusion chain $U_{n+1}\subseteq U_n$. Then there is a neighborhood $V$ of $f(x)$ and a sequence of points $(x_n)$ such that $x_n\in U_n$ and $f(x_n)\notin V$. Let $\phi_n:[\frac1{n+1},\frac1n]\to U_n$ be a path from $x_n$ to $x_{n+1}$, and take $\phi$ to be the union of these, extended to $[0,1]$ by setting $\phi(0)=x$. Now $\phi$ is continuous on $(0,1]$ by the pasting lemma, and it is continuous at $0$ because every neighborhood of $x$ contains some $U_n\supseteq \phi([0,\frac1n])$.
Thus $\phi$ is a path, so $f\circ\phi$ is continuous, and then there is an $n$ such that $f\circ \phi([0,\frac1n])\subseteq V$, which is a contradiction because $f\circ \phi(\frac1n)=f(x_n)\notin V$.
Is it necessary to assume first-countability here? If it helps, the application I want to use this theorem for is the claim that a covering map $p:Y\to Z$ lifts paths $f:X\to Z$ uniquely (given a basepoint $p(y_0)=f(x_0)$) to $\tilde f:X\to Y$ when $X$ is simply connected and locally connected. If I define $\tilde f$ such that $\tilde f(x)=\tilde \phi(1)$ for any path $\phi:[0,1]\to X$ from $x_0$ to $x$ (I have already shown that liftings exist for $X=[0,1]$ and $X=[0,1]^2$), then I can show that this defines a function which is "pathwise continuous" in the sense above, and I want to conclude that $\tilde f$ is the desired (continuous) lifting.