A bundle is a continuous map $p:Y\to X$ with a selected fiber $F$ so that for each $x\in X$, there is an open neighborhood $N_x$ of $x$ and a homeomorphism $p^{-1}N_x\cong N_x\times F$ such that $$(p^{-1}N_x\xrightarrow{\cong} N_x\times F\xrightarrow{\pi_1} N_x)=(p^{-1}N_x\xrightarrow{p}N_x).$$ It follows that $F\cong p^{-1}(x)$. Now let $p$ be a bundle with fiber $F$, and $f:A\to X$ a continuous map, so we can form the pullback of topological spaces
$$ \newcommand{\ra}[1]{\!\!\!\!\!\!\!\!\!\!\!\!\xrightarrow{\quad#1\quad}\!\!\!\!\!\!\!\!} \newcommand{\da}[1]{\left\downarrow{\scriptstyle#1}\vphantom{\displaystyle\int_0^1}\right.} % \begin{array}{llllllllllll} Z&\ra{}&Y\\ \da{p'}&&\da{p}\\ A&\ra{f}&X \end{array}. $$ How would I prove that $p'$ is again a bundle with fiber $F$?
I assume that we say let $a\in A$. Since $p$ is a bundle, there exists an open neighborhood $N_{f(a)}$ of $f(a)$ and a homeomorphism $p^{-1}N_{f(a)}\cong N_{f(a)}\times F$ such that $$(p^{-1}N_{f(a)}\xrightarrow{\cong} N_{f(a)}\times F\xrightarrow{\pi_1} N_{f(a)})=(p^{-1}N_{f(a)}\xrightarrow{p}N_{f(a)}).$$ Then we look at $f^{-1}(N_{f(a)})$ which is open since $f$ is continuous and contains $a$. I have no idea how to proceed from here. I am assuming that the open neighborhood of $A$ we are looking for is in fact $f^{-1}(N_{f(a)})$.
You have two completely separate questions going on here. I'll address the first. You should ask the other in a different question, though you should first look around MSE to see if something like this answers your question about homotopy invariance.
Pullback is a bundle
To prove that the pullback is again a fiber bundle with the same fiber, we make the following observations.
(1) The pullback of a trivial bundle is (canonically) trivial. If $B=Y\times F$, and $f:X\to Y$, then $$f^*B = X\times_Y (Y\times F)\simeq (X\times_Y Y)\times F \simeq X\times F.$$
(2) We can reinterpret the fiber bundle condition that every $x\in X$ has a neighborhood $U_x$ with $p^{-1}(U_x) \cong U_x\times F$ as saying that every $x\in X$ has a neighborhood $U_x$ such that the pullback $U_x\times_X B$ is trivializable.
(3) Then if $x\in X$, $y=f(x)\in Y$, $U_y\subseteq Y$ is an open neighborhood of $y$ on which $B$ is trivializable. Let $V_x = f^{-1}(U_y)$. Then (omitting associativity isos) $$ \begin{align} V_x\times_X X\times_Y B &\simeq V_x\times_Y B \\ &\simeq V_x\times_{U_y} U_y \times_Y B \\ &\cong V_x\times_{U_y} U_y \times F \\ &\simeq V_x \times F \\ \end{align} $$ The first iso is the natural unit iso, the second iso is again the unit, the third is by a choice of trivialization, and the fourth is the natural iso of observation (1). This completes the proof.