let $\mu$ be a probability measure in a separable hilbert space. Then for all $x\in H$, for all $\varepsilon >0$, there exists an $r>0$, such that $\mu(B_r(x))>1-\varepsilon$. $B_r(x)$ denotes the closed ball with radius $r$.
Proof: Fix a countable Basis $B_n$ of $H$. Define inductively $E_n=B_n\setminus \cup_{k<n}(B_k)$. $E_n$ are pairwise disjoint and cover $H$, hence $\mu(\cup_n E_n)=1$. By continuity from below, there exists an $N$, with $$\mu(\cup^N E_n)>1-\varepsilon$$ Take $r=\sup \text{dist}_{e\in \cup^NE_n}(x,e)$.
Question: is this proof correct? is that just the definition of inner regular measure? is every Borel probability measure inner regular in Hilbert space?
Your proof is incorrect and also a lot more complicated than necessary. Note that if $B = \{B_n: n \in \mathbb{N}\}$ is an orthonormal basis of $H$, as indicated in the comments, then $B$ is a subset of the unit ball of $H$. So just consider $\mu = \delta_{2B_1}$ to see that $\mu(\bigcup_n E_n) = \mu(B) = 0 < 1$.
A simple proof of this fact notices that $H = \bigcup_{r \in \mathbb{N}} B_r(x)$ and so $$1 = \mu(H) = \lim_{N \to \infty} \mu(\bigcup_{j = 1}^N B_j(x)) = \lim_{N \to \infty} \mu(B_N(x))$$ and so for $N$ large enough $\mu(B_N(x)) > 1- \varepsilon$.
To answer your question about inner regularity, I need to fix first what I will mean by inner regular since different sources I've seen differ.
I will use the definitions found in Bogachev's "measure theory". There, a Borel measure $\mu$ is called inner regular if for every Borel set $A$ we have $$\mu(A)=\sup\{\mu(K): K\subseteq A, K \text{ is closed}\}$$ and is called tight if $$\mu(A) = \sup\{\mu(K): K \subseteq A, K \text{ is compact}\}.$$ Note that with this definition, any tight measure is inner regular. However, the property stated is different from inner regularity because inner regularity tells you nothing when you take $A = H$ since $H$ is closed in itself.
Finally, using Bogachev's definitions it is true that every finite Borel measure on a metric space is regular but not every such measure is tight (see e.g. example 7.1.6 and theorem 7.1.7 in the second volume of Bogachev's measure theory).