When does $(X\times Y,\Vert\cdot\Vert)$ have the product topology? Let $(X,\Vert\cdot\Vert_X)$ and $(Y,\Vert\cdot\Vert_Y)$ be normed vector spaces and let $\Vert\cdot\Vert$ be a norm on $X\times Y$ such that $\Vert(x,0)\Vert=\Vert x\Vert_X$ and $\Vert(0,y)\Vert=\Vert y\Vert_Y$ for all $x\in X$ and $y\in Y$. Is the metric topology induced by this norm necessarily the product topology on $X\times Y$?
It suffices to show that this norm and the norm $\Vert(x,y)\Vert_1:=\Vert x\Vert_X+\Vert y\Vert_y$ are equivalent since $\Vert\cdot\Vert_1$ induces the product topology, but I was only able to show $\Vert\cdot \Vert$ is continuous with respect to $\Vert\cdot\Vert_1$, not conversely. This shows only that the metric topology is no finer than the product topology, not that they are equal. So, I suspect there may be a counterexample.
 A: Your suspicion is correct: the topology on $(X\times Y,\lVert \cdot\rVert)$ may be strictly coarser than the product topology. Namely, consider a Banach space $(W,\lVert \cdot\rVert)$, a proper dense subspace $X$ and a subspace $Y$ such that $X\oplus Y=W$. Then consider $(X\times Y,\lVert \cdot\rVert)$ with the norm of $W$ (which is effectively $W$ under the identification $(x,y)=x+y$) versus the product space $(X,\left.\lVert \cdot\rVert\right\rvert_X)\times(Y,\left.\lVert \cdot\rVert\right\rvert_Y)$. $\lVert(x,0)\rVert=\lVert x\rVert_X$ and $\lVert(0,y)\rVert=\lVert y\rVert_Y$, but the topology is not the product topology because $X\times\{0\}$ is closed in the product, but not in $W$. In fact, since the product space has a non-Banach norm, it can't be homeomorphic to a Banach space at all.
More generally, with the same assumptions as your question plus $\lVert\cdot\rVert$ Banach, the topologies are the same if and only if both $X$ and $Y$ are Banach.
Added: It should be noted that $W$ being Banach is superfluous for the purpose of making a counterexample. The same can be done with any normed space that has a non-closed vector subspace. As a side remark, in this generality the product space may still be homeomorphic to $W$: see here an example which exploits the fact that pre-Hilbert spaces of algebraic dimension $\aleph_0$ are linearly isometric.
