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I am following the terminology established in the book by John M. Lee. For a family $(X_\alpha)_{\alpha \in A}$ of topological spaces let $\prod_{\alpha \in A} X_\alpha$ be equipped with the usual product topology (not the box topology), i.e. the topology generated by the basis consisting of elements of the form $$\prod_{\alpha \in A}U_\alpha$$ where $U_\alpha$ is open in $X_\alpha$ and $U_\alpha = X_\alpha$ for all but finitely many $\alpha \in A$. Then we have the following theorem:

Theorem (Characteristic Property of Infinite Product Spaces). For any topological space $Y$, a map $f: Y \to \prod_{\alpha \in A}X_\alpha$ is continuous if and only if each of its component functions $f_\alpha = \pi_\alpha \circ f$ is continuous. The product topology is the unique topology on $\prod_{\alpha \in A}X_\alpha$ that satisfies this property.

From this theorem we easily deduce that each coordinate function $\pi_\alpha$ is continuous. Now I have read a few times that the product topology is also the smallest topology for which this holds (that each $\pi_\alpha$ is continuous). How do I see this?

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If we have a topology $\mathcal{T}$ for which each of the $\pi_\alpha$ is continuous, this means that for every $\alpha$, and every open subset $O$ of $X_\alpha$ we must have that $\pi_\alpha^{-1}[O] \in \mathcal{T}$. The finite intersections of sets of this form, say of $\pi_{\alpha_1}[O_1],\ldots \pi_{\alpha_n}^{-1}[O_n]$ is exactly the set $\prod_\alpha U_\alpha$ where $U_{\alpha_i} = O_{\alpha_i}$ for $i=1,\ldots,n$, and $U_\alpha = X_\alpha$ otherwis, is then also in $\mathcal{T}$, topologies being closed under finite intersections. So $\mathcal{T}$ contains the standard base for the product topology, and so $\mathcal{T}_{\text{prod}} \subseteq \mathcal{T}$. This establishes the minimality of the product topology w.r.t. the property that it makes all projections continuous.

The characteristic property is then also clear: suppose $f: Y \rightarrow \prod_\alpha X_\alpha$ is continuous, then all compositions with projections are necessarily continuous (as continuity is preserved by composition). But if we just assume that $\pi_\alpha \circ f$ is continuous for all $\alpha$, then take any basic open subset $B$ of $\prod_\alpha X_\alpha$ ,this is such an essentially finite box, and this we can write (just as above) as $B = \cap_i^n \pi_{\alpha_i}^{-1}[U_i]$ for some finite subset $\{\alpha_1,\ldots \alpha_n\}$ of $A$.

But then: $$f^{-1}[B] = f^{-1}[\cap_i^n \pi_{\alpha_i}^{-1}[U_i]] = \cap_{i=1}^n f^{-1}[\pi_{\alpha_i}^{-1}[U_i]] = \cap_{i=1}^n (\pi_{\alpha_i} \circ f)^{-1}[U_i]$$

which is a finite intersection of open subsets of $Y$ as all $\pi_{\alpha_i} \circ f$ are continuous and the $U_i$ are all open.

So the inverse images of basic open subsets of the product topology are open, so $f$ is continuous. ($f^{-1}$ preserves unions.)

Now think about why the product topology is the only such topology on the product.

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    Thanks for this wonderfull answer! It fixed my notational issues. If there would be another topology satisfying the characteristic property, we could consider $Y$ as the product space with this other topology and the identity mapping. Now the role can be exchanged which means the topologies coincide. I am a bit unhappy about this since the proof of the characteristic property uses the explicit basis elements of the product topology. – 2017-01-20
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    @TheGeekGreek Glad you like it.. You use the characteristic property twice (once for the product, once for the other candidate) to show the identity indeed is a homeomorphism, which is what's needed. The proof of the characteristic property should of course use the specific way the product is constructed. – 2017-01-20
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    Thanks again...it is like this existence and uniqueness proofs standard in algebra (e.g. tensor products). – 2017-01-20
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    @TheGeekGreek They're standard category theory arguments for the essential uniqueness of limits. – 2017-01-21