How to check F:AxI->B is continuous A and B are topological spaces.Let f and f' are continuous maps from A to B and homotopic.Then we need F:AxI->B,continuous,where F(s,0)=f(s) and F(s,1)=f'(s).
Now my question is if we want to explicitly construct some homotopy F between f and f' how should we try to prove that F is actually continuous?(i'm sorry if the question is not clear)
 A: Well, before getting to homotopy you should have been taking a general topology semester, haven't you? -And topology is about continuity, so hopefully you should have plenty of resources to prove that a certain map is continuous.
Nevertheless, if you're taking a standard topology course, when talking about homotopy, things are slightly easier because, in the usual examples and exercises, $A$ and $B$ are subsets of some $\mathbb{R}^n$ and there you just apply your knowledge about continuous mappings you got from a calculus course.
Even better: I would say that 90% of the homotopies you're going to encounter are variations of the straight line homotopy. That is, if $f(a)$ and $f'(a)$ are points in some $\mathbb{R}^n$, you can always join them with the segment $(1-t)f(a) + tf'(a)$ with $t\in [0,1]$. And this formula gives you a homotopy $F(a,t)$ from $f$ to $f'$ which is obviously continuous (i.e., standard calculus).
Finally, the only problem could be that $B \subset \mathbb{R}^n$ might not be a convex set. Hence, some of these straight lines upon which you've built your homotopy could leave $B$ at some points. And this is, perhaps, the only difficulty you're going to find with homotopies in a first course.
Example. To prove that $[0,1]$ is contractible, you need a homotopy from the identity of this interval to a constant map. Here it is one such a homotopy. Define 
$$
F : [0,1] \times [0,1] \longrightarrow [0,1] 
$$
as
$$
F(a,t) = (1-t)\ \mathbf{id} (a) + t\ \mathbf{constant \ map \ at \ 0}= (1-t)a \ .
$$
