= Solution
Because $V$ is measurable for the <cylinder sigma-algebra>, membership in $V$ depends on only countably many coordinates $T_0=\{t_1,t_2,\ldots\}$. Its image in $\mathbb R^{T_0}$ is a measurable linear subspace $V_0$, and
$$
\{X\in V\}=\{(X(t_j))_{j\geq1}\in V_0\}.
$$
By successively applying the finite-dimensional Gaussian regression formula, realize this Gaussian sequence as a lower-triangular linear transform of independent standard normal variables:
$$
(X(t_j))_{j\geq1}=\sum_{k\geq1}g_k a_k,
$$
where every coordinate of the sum contains only finitely many terms. If some deterministic column $a_k$ does not belong to $V_0$, then, after conditioning on every $g_j$ except $g_k$, at most one value of $g_k$ can put the sum in $V_0$. The continuous normal distribution gives probability zero. If every $a_k$ belongs to $V_0$, changing finitely many $g_k$ does not change the membership event. It is then a <tail event>, and the <Kolmogorov zero-one law> gives probability zero or one. This proves the <Gaussian zero-one law for measurable linear subspaces>.
Now let $X(t)=\sqrt t,g_t$ for independent standard normal variables $g_t$. Define
$$
V=\left\{x\in\mathbb R^{\mathbb N}:\frac{x_t}{t}\longrightarrow0\right\},
\qquad
W=\ell^2.
$$
Both are cylinder-measurable infinite-dimensional linear subspaces. For every $\varepsilon>0$,
$$
\sum_{t=1}^\infty\mathbb P(|g_t|>\varepsilon\sqrt t)<\infty,
$$
so the <Borel-Cantelli lemmas> imply $g_t/\sqrt t\to0$ almost surely and hence $\mathbb P(X\in V)=1$. On the other hand, $g_t^2\geq1$ infinitely often almost surely, again by Borel-Cantelli, so
$$
\sum_{t=1}^\infty|X(t)|^2=\sum_{t=1}^\infty t g_t^2=\infty
$$
almost surely. Therefore $\mathbb P(X\in W)=0$.
Solved by gpt-5.6-sol high.
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