I would like to prove (if possible, otherwise find a counterexample for) the following lemma:

Let $(X,\|\cdot \|_X)$ be a separable Banach space. Additionally, we have a centred Gaussian measure $\mu$ on $X$, with Cameron--Martin space $(E, |\cdot |_E)$.

We consider a sequence $x_n\in X$ with the following properties: $|x_n|_E\to \infty$ (wlog nondecreasing) and there exists a $C > 0$ such that $\inf_n \|x_n\|_X > C$.

Let $\delta_n$ be any nonincreasing sequence $\delta_n \to 0$ for $n\to \infty$. We define the open ball wrt $\|\cdot\|_X$: $$B_\delta (z) = \{x\in X: \|x-z\|_X < \delta\}.$$

Then (this is what I want to prove) $$ \lim_{n\to \infty} \frac{\mu(B_{\delta_n}(x_n))}{\mu(B_{\delta_n}(0))} = 0. \quad (*)$$

My thoughts so far:

For *fixed* $x\in X$, it is known (for example Bogachev, this is because the squared CM norm is the Onsager--Machlup functional, and it also follows from this thread: Probabilities of small balls with convergent center points under Gaussian measure) that
$$ \lim_{m\to \infty} \frac{\mu(B_{\delta_m}(x))}{\mu(B_{\delta_m}(0))} = e^{-|x|_E^2/2}. $$

Plugging a sequence $x_n$ as above in here:

$$\lim_{n\to\infty} \lim_{m\to \infty} \frac{\mu(B_{\delta_m}(x_n))}{\mu(B_{\delta_m}(0))} = \lim_{n\to \infty} e^{-|x_n|_E^2/2} = 0.$$

This means that (*) amounts to "applying both limits at once" instead of "sequentially". In my opinion, the missing piece here would be uniformity (with respect to $n$) of the limit

$$ \lim_{m\to \infty} \frac{\mu(B_{\delta_m}(x_n))}{\mu(B_{\delta_m}(0))}.$$

In other words, this expression should converge uniformly over $n$ by using $\inf_n \|x_n\|_X > C$ (which means that the balls $B_{\delta_m}(x_n)$ are uniformly bounded away from the origin). But I am unable to prove this. Maybe I am wrong and the statement is wrong after all?