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To find the product (for simplicity of two objects) in category of topological spaces $TOP$ one should use the universal property:

For any $Y \in Ob(TOP)$ and $\phi_{1,2}: Y \to X_{1,2} $ there is unique morphism (continuous map) $f: Y \to X_1 \times X_2 $ such that $\phi_{1,2} = \pi_{1,2} \circ f $. Now one deduces that (for instance) for any open $U \in X_1$ preimage $\pi_1^{-1}(U)$ should be open in $X_1 \times X_2$.

Now the thing which I don't understand is following: we are saing that preimages of open sets from $X_{1,2}$ are subbase of topology on the product. Why exactly subbase? How do we know that it isn't the base or something?

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    Generally the intersection of two such sets isn't of the form $\pi_k^{-1}(U)$. – 2017-01-03
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    @DanielFischer my answer below is your comment expanded a bit. – 2017-01-06

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In order for $\mathcal{B}$ to be a base, we need the property:

$$\forall B_1, B_2 \in \mathcal{B}: \forall x \in B_1 \cap B_2: \exists B_3 \in \mathcal{B}: x \in B_3 \subseteq B_1 \cap B_2$$

which follows from the rquirement that finite intersections of open sets are open, and so must be unions of base elements.

But consider $X_1 \times X_2$, $\emptyset \neq O_1 \subsetneq X_1$, $\emptyset \neq O_2 \subsetneq X_2$, both open. Then $B_1 = \pi_1^{-1}[O_1] = O_1 \times X_2$, $B_2 = \pi_2^{-1}[O_2] = X_1 \cap O_2$ are in the subbase for the product topology. But we can not find a set of this form inside $B_1 \cap B_2 = O_1 \times O_2$, so the collection of inverse images of open sets under the projections, does not satisfy the above base requirement.