I was wondering if that is true as well.
I think you can take the ideas from the answer of this question (Does a weaker form of the mean value property already imply harmonicity for continuous functions?) to solve it.
Edit: I was told to instead provide a detailed answer in this thread. Still the main ideas come from the aforementioned post.
I think either I am missing something in your original argument, or it has a mistake in it: Fixing $\varepsilon(x)$ so that the MVP with respect to $x$ holds in $B_{\varepsilon(x)}(x)$ does not immediately give you the MVP on the ball $B_{\varepsilon(x)}(x)$. At least not in the sense that for any point $y$ in this ball and any $R>0$ so that $B_{R}(y) \subseteq B_{\varepsilon(x)}(x)$ it holds
$$ u(y) = \frac{1}{n\omega_n R^{n-1}} \int_{\partial B_{R}(y)} u(y) dS(y). $$
In fact, if that were the case, you would directly get that $u$ is harmonic in $\Omega$, which directly implies that the MVP holds. The idea is that harmonicity, i.e. $\Delta u = 0$ (and $u$ twice differentiable) is a property that you can check locally.
The reason why you don't get the statement for all $y$, $R$ as above is that the LMVP is only applicable if $R \leq \varepsilon(y)$, which (as you said) could get arbitrarily small.
There is a non-continuous 'counterexample' that highlights this problem. Take $\Omega = \mathbb{R}^2$ and define $u$ by letting $u(x) := 1$ if $x_2>0$, $u(x) := 0$ if $x_2 = 0$ and $u(x) := -1$ if $x_2 < 0$.
This function satisfies the LMVP, but it is clearly not harmonic. The problem is that as you approach the $x_1$-axis the $\varepsilon(x)$ get arbitrarily small. A continuous counterexample does not exist, since (I believe) LMVP really does imply MVP, which I will try to sketch a proof for.
As we have seen, all we really need to show is that $u$ is harmonic. Take any $x\in \Omega$ and $r>0$ so that $\overline{B_r(x)} \subseteq \Omega$. Let $\tilde{u}$ be the solution to the Dirichlet problem
$$ \Delta \tilde{u} = 0 \quad \text{in $B_r(x)$}, $$
$$ \tilde{u} = u \quad \text{on $\partial B_r(x)$}. $$
Since $u$ is continuous and $B_r(x)$ has a sufficiently smooth boundary, a solution exists. The function $\tilde{u}$ is harmonic and thus satisfies the MVP, so trivially also the LMVP. Thus the function $v = u - \tilde{u}$ (as a function on $\overline{B_r(x)}$) is continuous and satisfies the LMVP. Note also that because of the boundary conditions, $v$ vanishes on $\partial B_r(x)$.
Next we show that $v$ takes its maximum on $\partial B_r(x)$ (this is the maximum principle, I will take a proof from one of my lectures, I believe it is similar in Evans' book). Since $v$ is a continuous function on a compact domain, it clearly obtains its maximum $M$ somewhere. Let $B_r(x)$ denote the open ball and define
$$ \Sigma := \{ y \in B_r(x) \colon v(y) = M \}.$$
Since $v$ is continuous, $\Sigma$ is relatively closed in $B_r(x)$. For any $y$ in $\Sigma$, choose $\varepsilon(y)$ as in the statement of the LMVP and let $\varepsilon' \leq \varepsilon(y)$ be so that $B_{\varepsilon'}(y) \subseteq B_r(x)$. Then for any $\delta \in (0,\varepsilon')$:
$$ M = v(y) = \frac{1}{n\omega_n \delta^{n-1}} \int_{\partial B_{\delta}(y)} v(y) dS(y) \leq \frac{1}{n\omega_n \delta^{n-1}} \int_{\partial B_{\delta}(y)} M dS(y) = M, $$
So the two integrals in the middle are equal. Since $v$ is contnuous and $v \leq M$, this gives $v \equiv M$ on $\partial B_\delta(y)$. By the arbitrariness of $\delta$ we get $v \equiv M$ on $B_{\varepsilon'}(y)$, i.e. $B_{\varepsilon'}(y) \subseteq \Sigma$. This shows that $\Sigma$ is relatively open in $B_r(x)$.
Since $B_r(x)$ is connected, $\Sigma = \emptyset$ or $\Sigma = B_r(x)$, in both cases it attains its maximum on the boundary.
We know that $v$ vanishes on the boundary, so we get $v \leq 0$ on $B_r(x)$.
Repeating the same argument for the minimum yields that $v \geq 0$ on $B_r(x)$, so in summary we get that $v \equiv 0$. This means that $u$ and $\tilde{u}$ are identical on $B_r(x)$, so $u$ is harmonic on $B_r(x)$. This is precisely what we wanted to show.
