Path homotopy and separately continuous functions

This example should convince you that separately continuous is not the right definition for a homotopy.

Consider $S^1$ as the unit circle in the complex plane. Take the map $F : I \times I \to X$ given by $$F(s,t) = \begin{cases} \exp(i \pi t), & 0 \le t \le s \le 1 \\ \underbrace{\exp \left(-i \pi \frac{(s+1)t - 2s}{1-s}\right)}_{= exp \left(i \pi \frac{(s+1)t - 2s}{s-1}\right)}, & 0 \le s < t \le 1 \\\\ \end{cases}$$ $F$ is separately continuous, $F(0,t) = \exp(-i \pi t)$, $F(1,t) = \exp(i \pi t)$, $F(s,0)=1$, $F(s,1)=-1$. So this is a "homotopy" between a path from 1 to -1 which moves clockwise around the unit circle, and one which moves counterclockwise.

In other words, with your definition, $S^1$ would be simply connected! So using separate continuity would not give us a homotopy theory that fits our intuition.

Note that $F(s,t)$, as a function of $t$, is a path that moves counterclockwise along the unit circle at speed 1 for the first $s$ seconds, and then uses the remaining time to move clockwise all the way to -1. As $s \to 1$, the clockwise trip is squeezed into a very short time, until at $s=1$ it disappears completely.

The key point is that as $s \to 1$, we have $F(s,t) \to F(1,t)$ pointwise in $t$, but not uniformly in $t$, so the paths $F(s,t)$ and $F(1,t)$ are really not very similar looking. For $s$ close to 1, you can't really justify calling $F(s,t)$ a "continuous deformation" of $F(1,t)$.

One reason is that we want to be guaranteed that for any map $g: I \hookrightarrow I \times I$ with $g(0) = (0,0)$ and $g(1) = (1,1)$, the composition $F \circ g$ to be continuous. This guarantees that we can continuously deform portions of the path independently, and is one of the main properties used to prove that multiplication (path composition) in the fundamental group is associative.