The following theorem (or its corollary) implies negative answer to the original question.

Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ has cardinality $|R_x|\le w(X)$, where $w(X)$ stands for the weight of the topology of the metric space $(X,d)$.

Proof. Fix a base $\mathcal W$ of the topology of the space $X$ of cardinality $|\mathcal W|=w(X)$. For every $r\in R_x$ choose a point $x_r\in\bar B(x,r)\setminus cl(B(x,r))$ and find a basic neighborhood $W_r\in\mathcal W$ of $x_r$, which is disjoint with the open ball $B(x,r)$. Assuming that $|R_x|>w(X)=|\mathcal W|$, we can find two distinct real numbers $r<\rho$ in $R$ such that $W_r=W_\rho$. Then $x_r\in W_r=W_\rho$ and $x_r\in \bar B(x,r)\subset B(x,\rho)$. So, $x_r\in W_\rho\cap B(x,\rho)=\emptyset$, which is a desired contradiction showing that $|R_x|\le w(X)$.

Corollary. For any point $x$ of a locally compact (more generally, locally separable) metric space $(X,d)$ there exists $\varepsilon>0$ such that the set $\{r\in(0,\varepsilon):cl(B(x,r))\ne\bar B(x,r)\}$ is at most countable.

The following theorem (or its corollary) implies negative answer to the original question.

Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ has cardinality $|R_x|\le w(X)$, where $w(X)$ stands for the weight of the topology of the metric space $(X,d)$.

Proof. Fix a base $\mathcal W$ of the topology of the space $X$ of cardinality $|\mathcal W|=w(X)$. For every $r\in R_x$ choose a point $x_r\in\bar B(x,r)\setminus cl(B(x,r))$ and find a basic neighborhood $W_r\in\mathcal W$ of $x_r$, which is disjoint with the open ball $B(x,r)$. Assuming that $|R_x|>w(X)=|\mathcal W|$, we can find two distinct real numbers $r<\rho$ in $R_x$ such that $W_r=W_\rho$. Then $x_r\in W_r=W_\rho$ and $x_r\in \bar B(x,r)\subset B(x,\rho)$. So, $x_r\in W_\rho\cap B(x,\rho)=\emptyset$, which is a desired contradiction showing that $|R_x|\le w(X)$.

Corollary. For any point $x$ of a locally compact (more generally, locally separable) metric space $(X,d)$ there exists $\varepsilon>0$ such that the set $\{r\in(0,\varepsilon):cl(B(x,r))\ne\bar B(x,r)\}$ is at most countable.

The following theorem (or its corollary) implies negative answer to the original question.

Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ has cardinality $|R|\le w(X)$, where $w(X)$ stands for the weight of the topology of the metric space $(X,d)$.

Proof. Fix a base $\mathcal W$ of the topology of the space $X$ of cardinality $|\mathcal W|=w(X)$. For every $r\in R_x$ choose a point $x_r\in\bar B(x,r)\setminus cl(B(x,r))$ and find a basic neighborhood $W_r\in\mathcal W$ of $x_r$, which is disjoint with the open ball $B(x,r)$. Assuming that $|R_x|>w(X)=|\mathcal W|$, we can find two distinct real numbers $r<\rho$ in $R$ such that $W_r=W_\rho$. Then $x_r\in W_r=W_\rho$ and $x_r\in \bar B(x,r)\subset B(x,\rho)$. So, $x_r\in W_\rho\cap B(x,\rho)=\emptyset$, which is a desired contradiction showing that $|R_x|\le w(X)$.

Corollary. For any point $x$ of a locally compact (more generally, locally separable) metric space $(X,d)$ there exists $\varepsilon>0$ such that the set $\{r\in(0,\varepsilon):cl(B(x,r))\ne\bar B(x,r)\}$ is at most countable.

The following theorem (or its corollary) implies negative answer to the original question.

Theorem. For any point $x$ of a metric space $(X,d)$ the set $R_x:=\{r>0:cl(B(x,r))\ne \bar B(x,r)\}$ has cardinality $|R_x|\le w(X)$, where $w(X)$ stands for the weight of the topology of the metric space $(X,d)$.

Proof. Fix a base $\mathcal W$ of the topology of the space $X$ of cardinality $|\mathcal W|=w(X)$. For every $r\in R_x$ choose a point $x_r\in\bar B(x,r)\setminus cl(B(x,r))$ and find a basic neighborhood $W_r\in\mathcal W$ of $x_r$, which is disjoint with the open ball $B(x,r)$. Assuming that $|R_x|>w(X)=|\mathcal W|$, we can find two distinct real numbers $r<\rho$ in $R$ such that $W_r=W_\rho$. Then $x_r\in W_r=W_\rho$ and $x_r\in \bar B(x,r)\subset B(x,\rho)$. So, $x_r\in W_\rho\cap B(x,\rho)=\emptyset$, which is a desired contradiction showing that $|R_x|\le w(X)$.

Corollary. For any point $x$ of a locally compact (more generally, locally separable) metric space $(X,d)$ there exists $\varepsilon>0$ such that the set $\{r\in(0,\varepsilon):cl(B(x,r))\ne\bar B(x,r)\}$ is at most countable.