Proof:

Let $p_1$ and $p_2$ be the usual projection maps $X\times Y\longrightarrow X$ and $X\times Y\longrightarrow Y$ respectively and $\gamma$ be a loop in $X\times Y$ based at $(x_0, y_0)$. Then $p_1\circ \gamma$ and $p_2\circ \gamma$ are loops in $X$ and $Y$ based at $x_0$ and $y_0$ respectively. The map

$$\phi:\pi_1(X\times Y, (x_0, y_0)) \longrightarrow \pi_1(X, x_0)\times \pi_1(Y, y_0)$$

$$\phantom{}[\gamma] \mapsto ([p_1\circ \gamma], [p_2\circ \gamma]) \phantom{XXX}$$

is well-defined and easily seen to be a surjective group homomorphism. Injectivity is also easy to check. Well, suppose that $[\gamma]$ is in the kernel of $\phi$ then $p_1\circ \gamma$ and $p_2\circ \gamma$ are homotopic to the constant loops $\varepsilon_{x_0}$ and $\varepsilon_{y_0}$ respectively via homotopies $F_1$ and $F_2$. That is to say there exists continuous maps $F_1:I^2 \longrightarrow X$ and $F_2:I^2 \longrightarrow Y$ such that

$$F_1(0, t) = p_1\circ\gamma, F_1(1, t) = \varepsilon_{x_0}, F_2(0, t) = p_2\circ\gamma, F_2(1, t) = \varepsilon_{y_0}.$$

and $F_1(s, 0) = F_1(s, 1) = x_0$, $F_2(s, 0) = F_2(s, 1) = y_0$ for all $s\in [0,1]$. Putting these together we get a continuous map $F_1\times F_2:I^2\longrightarrow X\times Y$ namely

$$(s, t)\mapsto (F_1(s, t), F_2(s, t))$$

which is a homotopy between $\gamma$ and the constant loop at $(x_0, y_0)$ proving that the kernel is trivial.