Let $X$ be a Hilbert space and $A\subset X$ be a compact subset. Assume that $T$ is the tangent space to $A$ at $a \in A$, in the sense that if $\phi :(-2,2) \to A$ is smooth with $\phi(0)=a$, then $\phi'(0) \in T$ (and conversely, any element of $T$ can be represented as $\phi'(0)$ for some $\phi$ satisfying the previous properties), say that $T \subset X$ is a closed subspace. Fix $x \in X$. Can we "linearize" the distance function (maybe under additional assumptions, or with less assumptions) as follows:
$$d(x,A) = d(x-a,T) + O(\|x-a\|^2)$$
Linearize here means that instead of calculating distance from the set $A$ we approximate it with distance from the subspace $a+T$. This probably should reduce to usual Taylor's formula on $\mathbb{R}$, but I have difficulty in seeing how. Using the compactness and closedness assumptions we know that the distances must be attained, say at $a_* \in A$ and at $t_* \in T$, so we should prove
$$I:=\|x-a_*\| - \|x-a-t_*\| = O(\|x-a\|^2)$$
By triangle inequality we can bound the above difference by $\|a_*-a-t_*\|$, now we can choose $\phi$ with $\phi'(0)=t_*$ and we can probably also assume that $\phi(1) = a_*$, so that
$$I\leq \|\phi(1) - \phi(0) - \phi'(0)\|$$
and I don't know how to bound this with $\|x-a\|^2$.