No, this need not be true, unless I am missing some assumption implicit in your use of the phrase "the induced norm" (emphasis added). Here is an example where $X=Y=\mathbb{R}$. All norms on $X\times Y=\mathbb{R}^2$ are equivalent. The norm $\|(x,y)\|^2=2x^2-2xy+y^2=x^2+(x-y)^2$ does not satisfy your property because, for example $\|(x,x)\|=|x|$ while $\|(x,0)\|=\sqrt{2}|x|$. (This norm actually comes from an inner product on $\mathbb{R}^2$.)
The following was a result of misreading the question. I'll leave it here, for now at least, if for no other reason than leaving Arturo's correction comprehensible:
What is true is that the projection map $(x,y)\mapsto x$ is continuous. This implies as a special case the following (which turns out to actually be equivalent to continuity):
For all $\epsilon>0$, there is a $\delta>0$ such that for all $x\in X$ and $y\in Y$, $\|(x,y)\|<\delta$ implies $\|x\|<\epsilon$.
In fact, the projection is Lipschitz continuous, and $\delta$ can be taken to be $\epsilon$ divided by the Lipschitz constant. For the reason Qiaochu Yuan gave in a comment on Ross Millikan's answer, this is the best you could hope for.