Prove that $\coprod_{i \in I} X_i$ metrizable iff $\forall i \in I: X_i$ metrizable
My attempt:
"$\implies$" If $\coprod X_i$ is metrizable, then also the subspace $X_i \times \{i\}$ for each $i \in I$, and because this space is homeomorph with $X_i$, it follows that $X_i$ is metrizable as well.
$"\impliedby"$ Suppose $\mathcal{T}_i$ is induced by a metric $d_i$ for every $i \in I$, which we can assume to be less than $1$. (Indeed, if $d$ is a metric, then the function $d \land 1$ is a metric too that induces the same topology).
Define $d: \coprod X_i \times \coprod X_i \to \mathbb{R}^+: ((x,i),(y,j)) \mapsto \begin{cases} d_i(x,y) \quad i = j \\ 1 \quad i \neq j\end{cases}$
We claim that $\coprod \mathcal{T}_i = \mathcal{T}_d$
"$\subseteq$" Sufficient: $\forall i \in I: B_i \in \mathcal{T}_i: B_i \times \{i\} \in \mathcal{T}_d$
Hence let $(x,i) \in B_i \times \{i\}$. Then $x \in B_i \in \mathcal{T}_{d_i}$, so we can find $\epsilon \in (0,1): B_{d_i}(x, \epsilon) \subseteq B_i$. Then $B_d((x,i),\epsilon) = B_d(x,\epsilon) \times \{i\} \subseteq B_i \times \{i\}$, hence the inclusion follows.
"$\supseteq$" Let $\epsilon \in (0,1)$. Let $(x,i) \in \coprod X_i$.
Then $B_d((x,i), \epsilon) = B_{d_i}(x,\epsilon) \times \{i\} \in \coprod \mathcal{T}_i$. Because the balls with radius smaller than $1$ form a basis, this is sufficient.
This ends the proof.
Does this look correct?
To me it looks quite correct. Not much to add, really. Maybe a short proof that $d$ is actually a metric? (some cases for the triangle inequality, mostly).