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The above is from Hatcher's book.

I am curious exactly where is the hypothesis that $X$ and $Y$ are path-connected being used in the proof? I don't see it being invoked explicitly, nor can I see where it is implicitly used.

Also, a second question would be are there any counter-examples where the proposition does not hold if $X$ or $Y$ are not path-connected?

Thanks!

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    Typically, you consider $\pi_1(X)$ for path-connected spaces. Otherwise you should at least specify a *base point*, i.e. consider $\pi_1(X,x_0)$ for a special $x_0\in X$. – 2017-01-11
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    If $X$ is not path-connected, then $\pi_1(X, x_1)$ and $\pi_1(X, x_2)$ have nothing to do with each other when $x_1,x_2$ lie in different path components. So for such spaces, keeping the base point in the notation is essential. – 2017-01-11
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    I see... We can only write $\pi_1(X)$ (rather than $\pi_1(X,x_0)$) if the space is path-connected. Thanks. – 2017-01-11

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Epanding @G. Sassatelli's remark: The symbol $\pi_1(X)$ is shorthand for "a group isomorphic to $\pi_1(X, x_0)$ for any $x_0 \in X$, where $X$ is path connected so that all these groups are in fact isomorphic."

So path connectedness is really being used in making the statement of the theorem make sense.

You might look at $X = \{P\} \cup S^1$, and $Y = S^1$. The product of $X$ and $Y$ has two components -- a circle and a torus -- and it matters which one you work with.

An alternative formulation of the theorem would be something like this:

"For spaces $X'$ and $Y'$ containing points $x_0$ and $y_0$, let $X$ and $Y$ be the path-components containing $x_0$ and $y_0$ respectively. Then $\pi_1(X \times Y) = \pi_1(X) \times \pi_1(Y)$, and indeed, $\pi_1(X' \times Y', (x_0, y_0)) = \pi_1(X, x_0) \times \pi_1 (Y, y_0)$, but this latter equality is seldom useful in practice, as it ignores all other components of the product space."

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They are assumed to be path connected to make sure that the symbol $\pi_1(X)$ is well defined. Note however that the assumption is really unnecessary, since

$$\pi_1(X\times Y, (x, y))\simeq \pi_1(X, x)\times\pi_1(Y, y)$$

which indeed is a part of the proof. This even holds for homotopy groups of higher orders.

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There is confusion in the literature between a space and a space with base point. Grothendieck wrote to me in 1983:"... the choice of a base point, and the $0$-connectedness assumption, however innocuous they may seem at first sight, seem to me of a very essential nature. To make an analogy, it would be just impossible to work at ease with algebraic varieties, say, if sticking from the outset (as had been customary for a long time) to varieties which are supposed to be connected. Fixing one point, in this respect (which wouldn’t have occurred in the context of algebraic geometry) looks still worse, as far as limiting elbow-freedom goes!...." He was also keen on the use of the fundamental groupoid $\pi(X,C)$, or $\pi_1(X,C)$, on a set $C$ of base points, chosen according to the geometry of a situation.

The particular issue of products is dealt with in Topology and Groupoids by showing that the natural morphism $\pi(X \times Y ) \to \pi(X) \times \pi(Y)$ is an isomorphism, where $\pi(X)$ denotes the fundamental groupoid. What do you expect for $\pi(X \times Y, C\times D)$?

The general issue of using groupoid (and higher groupoid) structures in algebraic topology is discussed as part of this preprint Modelling and Computing Homotopy Types:I. In particular, $\pi_1(X,C)$ is useful for modelling and computing homotopy $1$-types.