For functions $f_1:X\to Y_1$ and $f_2:X\to Y_2$ define $f_1\nabla f_2:X\to Y_1\times Y_2$ by $(f_1\nabla f_2)(x)=(f_1(x),f_2(x))$.
Prove that $f_1\nabla f_2$ is continuous if and only if both $f_1$ and $f_2$ are continuous.
So, $f_1\nabla f_2$ is continuous then for each $x$ in $X$, and each neighborhood $V$ of $f_1\nabla f_2$, there is a neighborhood $U$ of $x$ such that $f_1\nabla f_2(U)$ contained in $V$. Let $x_1$ be a point in $X$ with $y_1=f_1(x_1)$ and $y_2=f_2(x_1)$.
Choose neighborhood $Vy_1$ and $Vy_2$ around $y_1$ and $y_2$ respectively. Now we have to find a neighborhood $Ux_1$ around $x_1$ such that $f_1\nabla f_2(x_1)$ contains $Vy_1$ and $Vy_2$.
Is this correct so far? If so, how should I proceed? Also, what is a good way to prove the opposite direction?