This should be a basic epsilon delta practice but I don't know how to prove it. Define $H: X\times[0,1]\to Y$ by $ f(x) = \begin{cases} F(x,2t) & t\in[0,\dfrac1 2] \\ G(x,2t-1) & t\in[\dfrac1 2, 1] \\ \end{cases} $
where F and G are continuous functions from $X\times[0,1]$ to $Y$ and $F(x,1)=G(x,0)$. I want to show that $H$ is continuous. It suffices to show that H is continuous at $t=\dfrac1 2$ since it is already continuous in $X\times[0,1]\setminus\{\frac1 2\}$. I wanted to state that F and G both converges to the same function at $t=\frac 1 2$ but it is not enough because it has to be continuous at every path around $(x,\frac1 2)$, I assume.
Well, just roll up your sleeves and do it.
Consider the point $(w,\frac 12)$ and let $\epsilon > 0$. As $F,G$ are continuous there are $\delta_F, \delta_G$ so that if $D((x,y) - (w,\frac 12)||< \delta_{F,G}$ then $||F(x,2y)-F(w,1)|| < \epsilon$ and $||G(x,2y-1)-G(w,0)|| < \epsilon$. Let $\delta = \min(\delta_F, \delta_G)$.
Now consider a point $(u,v)$ so that $||(u,v)-(w,\frac 12)|| < \delta$. If $v < \frac 12$ then $||F(u,2v)-F(w,1)||=|f(u)-f(w)| < \epsilon$. If $v> \frac 12$ then $||G(u,2v-1)-G(w,0)||=|f(u)-f(w)| < \epsilon$. If $v = \frac 12$ then $F(x,2v)=G(x,2v-1)$ and $||F(u,2v)-F(w,1)||=||G(u,2v-1)-G(w,0)||=|f(u)-f(w)|< \epsilon$.
And we are done.