If $V \times W$ with the product norm is complete, must $V$ and $W$ be complete?

Let $V,W$ be two normed vector spaces (over a field $K$). Then their product $V \times W$ with the norm $\|(x,y)\| = \|x\|_V + \|y\|_W$ is a normed space.

Using this norm it's easy to show that if $V,W$ are complete then so is $V \times W$. To see this, let the limit of the sequence $(x_n , y_n)$ be $(x,y) = (\lim x_n, \lim y_n)$. Then for $n$ large enough, both $\|x - x_n\|_V$ and $\|y - y_n\|_W$ are less than $\varepsilon / 2$ and hence $\|(x,y) - (x_n, y_n)\|< \varepsilon$.

The other direction does (probably) not hold. Can someone show me an example of a space $V \times W$ that is complete but either $V$ or $W$ (or both) are not?

If $V$ is not complete and $V\times W$ complete, take $\{v_n\}$ a Cauchy sequence which doesn't converge in $V$. Then $(v_n,0)$ is a Cauchy sequence in $V\times W$ and converges to $(v,w)\in V\times W$. We have $$\lVert (v_n,0)-(v,w)\rVert_{V\times W}=\lVert v_n-v\rVert+\lVert w\rVert\to 0$$ hence $v_n\to v$ in $V$, a contradiction.

Hence $V\times W$ is complete if and only if so are $V$ and $W$.

I believe that the spaces $V$ and $W$ must be complete whenever $V\times W$ is complete.

Closed subspace of a complete normed space is complete.

The space $V$ is isometrically isomorphic to the closed subspace $V\times\{0\}$ of $V\times W$.