Is the sum of two complete metrics complete? Let $X$ a space with two complete metrics $d_1$, $d_2$:
Is $d=d_1+d_2$ complete?

Let $(X,d_1)$ be complete.
Let $f\colon X\to X$ be a bijection and defined $d_2(x,y)=d_1(f(x),f(y))$. Then $(X,d_2)$ is also complete (and in fact $f\colon(X,d_2)\to (X,d_1)$ is an isometry).
If you pick $f$ wild enough, $(X,d_1+d_2)$ will not be complete.
For example, start with $X=\mathbb R$ and $d_1$ the standard metric. Let $$f(x)=\begin{cases}x&\text{if $x\in\mathbb Q$}\\x+1&\text{if $x\notin\mathbb Q$}\end{cases} $$
Now pick a rational sequence $(q_n)_n$ converging (in standard $\mathbb R$) to an irrational $\alpha$. Then this is a Cauchy sequence with respect to $d_1$, $d_2$, and $d_1+d_2$. With respect to $d_1$, it converges to $\alpha$. With respect to $d_2$, it converges to $\alpha-1$. With respect to $d_1+d_2$, it does not converge: For obvious reasons there is no rational limit; and the distance between any rational and any irrational is $> \frac12$.
