Let $Top$ be the category of the topological spaces with continuous functions as morphisms. We have arbitrary morphisms $f:X\rightarrow Z,g:Y\rightarrow Z\in Top$. Then we get the fibre product $X\times_ZY=\{(x,y)\in X\times Y\,|\, f(x)=g(y)\}$ and a commutative diagram $\require{AMScd}$ \begin{CD} X\times_ZY @>{pr_2}>> Y\\ @V{pr_1}VV @VV{g}V\\ X @>{f}>> Z. \end{CD}
Why is $pr_1$ is continuous (For all open sets $O\subseteq X$ is $pr_1^{-1}(O)$ open.) ?
With the answer of Papuseme I got the following prove: With $X\times_ZY=\{(x,y)\in X\times Y\,|\, f(x)=g(y)\}$ we form, with the $x$ coordinates, a subset $U\subseteq X$. We project from the fibre product into $U$ with $p_U:X\times_ZY\rightarrow U\,|\,(x,y)\mapsto x$. Let $O\subseteq U$ be open, then $p_U^{-1}(O)=O\times Y$. With the definition of the product topology, we know, that $O\times Y$ is one of the open sets in the basis of the product topology. Hence, $p_U$ is continuous. Now we use the inclusion map $i_U:U\hookrightarrow X\,|\,x\mapsto x$. Let $P\subseteq X$ be open, then $i_U^{-1}(P)=P\cap U$ and by the definition of the subspace topology, the set $P\cap U$ is open. Hence, $i_U$ is continuous. Now we get $pr_1=i_U\circ p_U$, hence $pr_1$ is continuous.
It is clear that the x coordinates of the fibered product form a subset of X. Project from the fibered product to that subset, and utilize the definition of the product topology. Then project from that subset into X, expanding range is continuous. Now composition of continuous functions is continuous.