Let's stick with a particular point on the interval $0 \times [-1, 1]$, say $p = (0, 0)$. Notice that there is no need to make your neighborhoods balls centered at $p$; we could consider open squares instead (they also form a basis for the topology of the plane). Let $U_\epsilon := (-\epsilon, \epsilon) \times (-\epsilon, \epsilon)$ be some open square centered at $p$ (where $\epsilon > 0$). Then $U_\epsilon \cap \overline{S}$ consists of $0 \times (-\epsilon, \epsilon)$ and the graph of the function $\sin(1/x)$ restricted to the domain $D_\epsilon:=\{x \in (0, \epsilon) : |\sin(1/x)| < \epsilon\}$. You should be picturing a bunch of very short curve segments which are almost vertical. We can choose $\epsilon$ small enough that $D_\epsilon$ does not contain any $x$ such that $\sin(1/x) = 1$.
Now let $V$ be some nonempty open subset of $U_\epsilon$ containing $p$. It contains $U_{\epsilon'}$ for some smaller $\epsilon' > 0$. Then there exists some $x_0 \in (0, \epsilon')$ such that $\sin(1/x_0) = 1$ and $(x_0, \infty) \cap D_{\epsilon'} \neq \emptyset$ (this should be easy to see; there is a sequence of such $x_0$ which converges to $0$). It follows that $D_{\epsilon'} = (D_{\epsilon'} \cap (0, x_0)) \cup (D_{\epsilon'} \cap (x_0, \infty))$, i.e. it is disconnected.
I claim that we can use this information to prove that $V \cap \overline{S}$ is disconnected. The idea is to look at the intersections of this set with $(-\infty, x_0) \times \mathbb{R}$ and with $(x_0, \infty) \times \mathbb{R}$. Note that neither of these intersections is empty since, in either case, we can take an appropriate value of $x \in D_{\epsilon'}$ and note that $(x, \sin(1/x)) \in V$. Secondly, these open sets do indeed cover $V \cap \overline{S}$ since $(x_0, 1)$ is the only point in $\overline{S}$ whose $x$-coordinate is $x_0$, and $V$ contains no point whose $y$-coordinate is $1$. So we conclude that $V \cap \overline{S}$ is disconnected.
Well, ultimately the issue is that you aren't using the definition of connectedness. You're using your own vague intuition about how it's supposed to behave, but that's not how connectedness is actually defined. If you want to understand why a space is connected or not, you have to actually use the definition.
The proof that the Knaster-Kuratowski fan is connected is fairly complicated, but here's a much simpler example that shows your intuition can't be right. Let $K$ be the Knaster-Kuratowski fan with its dispersion point removed, and consider the space $X=K\cup\{\infty\}$ with the topology that a set $U\subseteq X$ is open iff either $U=X$ or $U$ is an open subset of $K$. Then $X$ is connected, since the only open set containing $\infty$ is all of $X$. But if you remove $\infty$ from $X$, you're left with $K$ with its usual topology, which is totally disconnected.
As for why $K$ is totally disconnected, keep in mind that the definition of totally disconnected is not "any two points can be separated by clopen sets". Rather, it is "no subset with more than one point is connected". So even though you can't separate two points along the same line of $K$ by clopen subsets of $K$, they are still "disconnected" from each other in $K$ since there is no connected subset of $K$ that contains both of them.
Best Answer
This is not true in general. One needs to assume $X$ is at least $T_1$ in addition.
Ulli has already given a $T_0$ counterexample and a proof for Hausdorff spaces.
Let me add a proof in the case $X$ is only $T_1$.
In this case, suppose $x_0$ is a dispersion point of $X$, so $X\backslash \{x_0\}$ is totally disconnected. Suppose by contradiction $\gamma\colon [0,1]\to X$ is a nonconstant path in $X$. Then since $\{x_0\}$ is closed, $\gamma^{-1}(\{x_0\})$ is closed.
Since $\gamma^{-1}(\{x_0\})$ is closed, its complement is a disjoint union of open intervals, and by total disconnectedness of $X\backslash \{x_0\}$, $\gamma$ is constant on each complementary interval. But then the preimage of any point $x\in \gamma([0,1])\backslash \{x_0\}$ is open, hence not closed, (since $\gamma$ is nonconstant and $[0,1]$ is connected), contradicting the $T_1$ property of $X$.
Remark
We cannot replace $T_1$ by any weaker condition here, since if $X$ is not $T_1$, then for some distinct $x,y\in X$ we have $x\in \overline{\{y\}}$, at which point $x$ and $y$ can be connected by the path in Ulli's counter-example: $$\gamma(t)= \begin{cases} x & t=0\\ y & t\neq 0. \end{cases} $$ On the other hand, the existence of a dispersion point $x_0$ already implies $X\backslash \{x_0\}$ is $T_1$, so that verifying the $T_1$ condition in specific cases reduces to establishing that $x_0\notin \overline{\{x\}}$ and $x\notin \overline{\{x_0\}}$ for arbitrary $x\neq x_0$.