Prove that the fundamental group functor preserves products

Question

Is my proof correct?

Theorem: Let $(X, p)$ and $(Y,q)$ be pointed topological spaces. Then we have that $$\pi_1(X\times Y, (p,q))=\pi_1(X,p)\times\pi_1(Y,q).$$

Proof: Let $I=[0,1]$ and $\partial I=\{0,1\}$.

The idea is to prove $[(u,v)]=([u],[v])$ for arbitrary paths $u:I\to X$ and $v:I\to Y$.

In order to do so, we proceed as follows:

1. Let $u,u'$ be paths in $X$ and $v,v'$ paths in $Y$. Therefore $(u,v)$ and $(u',v')$ are paths in $X\times Y$. The assertion $(u,v) \simeq (u',v')$ Rel $\partial I$ means that there exists a homotopy $H_p:I^2\to X\times Y$ s.t. $H_p(t,s)=(u(t),v(t))=(u'(t),v'(t))$ $\forall s\in I$, $\forall t\in\partial I$.

2. $u\simeq u'$ Rel $\partial I$ and $v\simeq v'$ Rel $\partial I$ mean that there are homotopies $H_1:I^2\to X$ and $H_2:I^2\to Y$ s.t. $H_1(t,s)=u(t)=u'(t)$ $\forall s\in I$, $\forall t\in\partial I$, and $H_2(t',s')=v(t')=v'(t')$ $\forall s'\in I$, $\forall t'\in\partial I$.

3. We note that the map $S:I^2 \to I^2\times I^2$ s.t. $(t,s)\mapsto (t,s,t,s)$ is a diagonal map, in particular, it is continuous.

4. We can take the Cartesian product $\hat H_p := H_1\times H_2:I^2\times I^2\to X\times Y$ which is continuous since $H_1$ and $H_2$ are continuous.

5. We complete the proof by noting that $H_p=\hat H_p\circ S$.