Let us say that a topological space $X$ is spherically completely metrizable if the topology of $X$ is generated by a spherically complete metric.
Theorem. Every closed subspace $X$ of the countable product of locally compact metrizable spaces is spherically completely metrizable.
Proof. We lose no generality assuming that $X$ is a closed subspace of the countable power $L^\omega$ of some locally compact metrizable space $L$. By the paracompactness, the locally compact metrizable space $L$ is a topological sum $\bigcup_{\alpha\in\kappa}L_\alpha$ of clopen $\sigma$-compact subspaces. Each space $L_\alpha$ is locally compact and $\sigma$-compact, and hence its topology is generated by a metric $d_\alpha$ whose closed balls are compact. On the space $L$ consider the metric $d$ defined by
$$d(x,y)=\begin{cases}1&\mbox{if $x\in L_\alpha$ and $y\in L_\beta$ for distinct $\alpha,\beta\in\kappa$};\\
\min\{d_\alpha(x,y),1\}&\mbox{if $x,y\in L_\alpha$ for some $\alpha\in\kappa$}.
\end{cases}
$$
It follows that for every $c\in L$ and every $r<1$ the closed ball $B_L(c;r):=\{x\in L:d(x,c)\le r\}$ is compact and for every $r\ge 1$, $B_L(c;r)=L$.
On the countable power $L^\omega$ consider the complete metric $\rho$ defined by $$\rho((x_n)_{n\in\omega},(y_n)_{n\in\omega})=\max_{n\in\omega}\frac{d(x_n,y_n)}{2^n}.$$
We claim that the metric $\rho$ induces a spherically complete metric on the closed subspace $X\subseteq L^\omega$.
Let $(B_n)_{n\in\omega}$ be a sequence of nested closed balls in $X$. Let $(c_n)_{n\in\omega}$ and $(r_n)_{n\in\omega}$ be the sequences of the centers and radii of the balls $B_n$. Since every sequence of positive real numbers contains a monotone subsequence, we lose no generality assuming that the sequence $(r_n)_{n\in\omega}$ is monotone.
If the sequence $(r_n)_{n\in\omega}$ is increasing (i.e., $r_n\le r_{n+1}$ for all $n$), then for every $n\in\omega$ we have $c_n\in B_n\subseteq B_0$ and hence $\rho(c_n,c_0)\le r_0\le r_n$ and $c_0\in B(c_n;r_n)=B_n$.
So, we assume that $(r_n)_{n\in\omega}$ is strictly decreasing.
If $\inf_{n\in\omega}r_n=0$, then $\lim_{n\to\infty}r_n=0$ and the intersection $\bigcap_{n\in\omega}B_n$ is not empty by the completeness of the metric $\rho$.
So, we assume that $r:=\inf_{ n\in\omega}r_n>0$.
If $r\ge 1$, then every ball $B_n$ coincides with $X$ and hence $\bigcap_{n\in\omega}B_n=X\ne\emptyset$.
So, we assume that $r<1$. Let $m\in\omega$ be the largest number such that $2^m r<1$.
Since $\lim_{n\to\infty}r_n=r$, we can replace the sequence $(B_n)_{n\in\omega}$ by a suitable subsequence, and assume that $2^mr_0<1$. Then for every $k\in\omega$, the ball $B_L(c_0(k),2^mr_0)$ in $L$ is compact. For every $n\in\omega$, the inclusion $c_n\in B_n\subseteq B_0$ implies $\rho(c_n,c_0)\le r_0$ and hence $d(c_n(k),c_0(k))\le 2^kr_0$. Then for every $k\le m$, the sequence $(c_n(k))_{n\in\omega}$ is contained in the compact ball $B_L(c_0(k),2^mr_0)$ and hence has a convergent subsequence.
Replacing $(B_n)_{n\in\omega}$ by a suitable subsequence, we can assume that for every $k\le m$ the sequence $(c_n(k))_{n\in\omega}$ is convergent in $L$, and moreover $d(c_i(k),c_j(k))<r$ for all $i,j\in\omega$.
We claim that $\rho(c_0,c_n)\le r_n$ for every $n\in\omega$. This inequality will follow as soon as we check that $d(c_0(k),c_n(k))\le 2^kr_n$ for all $k\in\omega$.
If $k>m$, then $d(c_0(k),c_n(k))\le 1\le 2^{m+1}r<2^kr_n$ by the definition of the metric $d$.
If $k\le m$, then $d(c_0(k),c_n(k))<r\le 2^kr_n$ by the choice of the (sub)sequence $(B_i)_{i\in\omega}$.
Therefore, $c_0\in \bigcap_{n\in\omega}B_n$. $\quad\square$.
Since every Polish space is homeomorphic to a closed subspace of $\mathbb R^\omega$, Theorem implies
Corollary. Every Polish space is spherically completely metrizable.
The Theorem suggests the following
Question. Which metrizable spaces do embed into countable products of locally compact metrizable spaces?
Remark 1. The necessary condition of the embeddability of a topological space $X$ into the countable product of locally compact metrizable spaces is the separability of all quasicomponents of $X$. This condition implies that nonseparable connected metrizable spaces do not embed into countable products of locally compact metrizable spaces.
The following proposition answers the above Question.
Proposition. A topological space $X$ is homeomorphic to a closed subspace of the countable product of locally compact metrizable spaces if and only if $X$ is homeomorphic to a closed subspace of $\mathbb R^\omega\times\kappa^\omega$ for some cardinal $\kappa$ endowed with the discrete topology.
Proof. The "if" part of this characterization is trivial. To prove the "only if" part, assume that $X$ is homeomorphic to a closed subspace of the product $\prod_{n\in\omega}L_n$ of locally compact metrizable spaces $L_n$. By the paracompactness, every space $L_\alpha$ is a topological sum of locally compact $\sigma$-compact metrizable spaces and hence is a topological sum of Polish spaces. Since every Polish space is homeomorphic to a closed subspace of the space $\mathbb R^\omega$, for every $n\in\omega$ the locally compact metrizable space $L_n$ is homeomorphic to a closed subspace of $\mathbb R^\omega\times\kappa$ for some cardinal $\kappa$. Then $\prod_{n\in\omega}L_n$ is homeomorphic to a closed subspace of the space $(\mathbb R^\omega\times\kappa)^\omega$, which is homeomorphic to $\mathbb R^\omega\times\kappa^\omega$. $\quad\square$
Proposition and Theorem imply that every closed subspace of $\mathbb R^\omega\times\kappa^\omega$ is spherically completely metrizable.
Problem. Let $\kappa$ be a cardinal. Is every closed metrizable subspace of the space $[0,1]^\kappa\times\kappa^\omega$ spherically completely metrizable?
Remark 2. For every cardinal $\kappa$, every closed metrizable subspace of the space $[0,1]^\kappa\times\kappa^\omega$ is completely metrizable. On the other hand, closed metrizable subspaces of the space $\mathbb R^\kappa$ are realcompact but needs not be completely metrizable (by Theorem 3.11.12 in Engelking's "General Topology", every Lindelof space is realcompact; in particluar, every metrizable separable space is realcompact).
