The universal property of product topological space is as follows:
Let $Z,~Y_\alpha(\alpha\in A)$ be topological spaces. For every $g_\alpha: Z\to Y_\alpha$ continuous, there exists exactly one $g:Z\to \prod_{\alpha\in A}Y_\alpha$ such that $p_\alpha\circ g=g_\alpha$ for every $\alpha\in A$.
And the definition of universal/couniversal object in category theory is:

So applying such definition in the previous context, is it to say that $\prod_{\alpha\in A}Y_\alpha$ is a couniversal object in the topological space category?
No, it is not. A one-point space would be terminal: given any space $X$, there is one and only one continuous map $X \to \{*\}$, namely the constant map. There will typically be MANY maps of the form $f: Z \to \prod_\alpha Y_\alpha$, given only $Z$ and the $Y_\alpha$. The product is the unique "receiver" of a space $Z$ subject to satisfying the conditions you state, given the input maps $Z \to Y_\alpha$. This is very different from the terminal definition.
For example, how many continuous maps are there $f: \mathbb{R} \to \mathbb{R} \times \mathbb{R}$? Clearly, infinitely many. So, the euclidean plane is not a terminal object. Now, how many maps are there $f: \mathbb{R} \to \mathbb{R} \times \mathbb{R}$ that are continuous and also satisfy $f(x) = (\sin x, x^2)$? Clearly only one. That's the difference: the second question contains much more data/constraints on $f$. This is the universal property of the product of spaces.
To be fair, products and terminal objects are special cases of "limits" in category theory. Disjoint unions and initial objects are special cases of "colimits." So yes, terminal objects and products are of the same "flavor" but the input data is very different.