The aim of this problem is to guess whether the affirmation is true or not. I have proved the oposite implication by using the fact that if $X$ is contractible, then $X \times Y$ has the same homotopy type as $Y$.
But for this implication I have some doubts. I tried to prove it true by creating the homotopy $H:Id_{X \times Y} \simeq f_a$, taking $a = (a_1, a_2)$, $a_1 \in X, a_2 \in Y$ as the constant that satisfies $X \times Y \simeq \{a\}$ and $f_a$ the constant map, using that $X \times Y$ is contractible. From that, I create the following diagram:
$\require{AMScd}$
\begin{CD}
X\times Y\times \mathbb{I} @>{H}>> X \times Y\\
@V p_1 \times p_3 VV @VV p'_1 V\\
X \times \mathbb{I} @>{H_1}>> X
\end{CD}
which satisfies $$H_1(x,0) = p'_1(H(x,y,0)) = p'_1(Id_{X \times Y}) = Id_X$$ $$H_1(x,1) = p'_1(H(x,y,1)) = p'_1(f_a) = f_{a_1}.$$
So, the last step is to prove that $H_1$ is continous, proving that we obtain that $X$ is contractible. Of course, if it is true then the process would be the same to prove that $Y$ is contractible. But in this step is when my doubts appear, because I don't know if I can say it directly from diagram without making any wrong assumption.
Best Answer
The problem with your proof is that you have not defined $H_1$. You only say that you "create" the diagram, but there is no indication on how to actually produce the missing map. If the map is not defined, it's going to be difficult to prove that it's continuous...
But here you can define $H_1$ very simply and it will be obvious that it's continuous. Simply take $$H_1(x,t) = p_X(H(x,y_0,t))$$ where $p_X : X \times Y \to X$ is the projection and $H : X \times Y \times [0,1] \to X \times Y$ is a homotopy between $\operatorname{id}_{X \times Y}$ and the constant map equal to $(x_0, y_0) \in X \times Y$. It's clear that it's continuous (it's the composition of continuous maps), and the following properties are satisfied: $$H_1(x,0) = p_X(H(x,y_0,0)) = p_X(x,y_0) = x,$$ $$H_1(x,1) = p_X(H(x,y_0,1)) = p_X(x_0,y_0) = x_0.$$ In other words, $H_1$ is a homotopy between $\operatorname{id}_X$ and the constant map equal to $x_0$. So $X$ is contractible.