Direct sum of metrizable spaces. I managed to prove that an arbitrary direct sum of metrizable spaces is again metrizable. However, I used the theorem that says that a hausdorff regular space is metrizable if and only if there existd a $\sigma$-locally finite base. I was planning to show my proof for some friends, but they never saw the proof of this equivalence, therefore I was trying to create a simpler proof.
I am also trying to show that every direct sum of completely metrizable spaces is again completely metrizable and I am having no success.

SKETCH: Suppose that you have metric spaces $\langle X_i,d_i\rangle$ for $i\in I$. First replace $d_i$ be $d_i'$, where
$$d_i'(x,y)=\min\{d_i(x,y),1\}$$
for $x,y\in X_i$. Observe that $d_i'$ generates the same topology as $d_i$ and is complete if $d_i$ is complete. Now define a metric $d$ on $\bigsqcup_{i\in I}X_i$ by
$$d(x,y)=\begin{cases}
d_i'(x,y),&\text{if }x,y\in X_i\text{ for some }i\in I\\
1,&\text{if }x\in X_i,y\in X_j,\text{ and }i\ne j\;.
\end{cases}$$
Show that $d$ is a metric that generates the right topology and that is complete if all of the $d_i$ are complete.
