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 $x\neq x_0$.
Here is a proof:
$X$ is biconnected, finite, $n := |X| \ge 4$.
Let $\le$ be the specialization preorder on $X$, i.e.
$x \le y \Leftrightarrow x \in \overline{\{y\}}$.
every chain $C$ in $X$ has at most 2 elements.
[Assume $C = \{a, b, c\} $ are pairwise distinct, $a \le b \le c$. Then $C = \overline{\{c\}}$: "$\subset$" is clear.
"$\supset$": Let $x \in \overline{\{c\}}$ with $x\ne c$. Then $\{x,c\}$ and $\{a,b\}$ are connected, hence $x=a$ or $x=b$ (see below). Analogously, $C = \{x \in X: a <= x\}$. Hence C is closed and open, hence $C= X$. Contradiction.]
X is T0, i.e. $\le$ is a partial order.
[Assume not, then there are distinct $a, b \in X$, such that $a \le b \le a$. By 1. we have
$\overline{\{a\}} = \overline{\{b\}} = \{a, b\}$,
and for $x \in X$: if $a \in \overline{\{x\}}$, then
$x \in \{a, b\}$.
Hence for $x \in X \setminus \{a, b\}$:
$\overline{\{a\}} \cap \overline{\{x\}} = \emptyset$ and
$\overline{\{x\}} = \{x\}$, since X is biconnected.
Since X is finite, we have $X \setminus \{a, b\}$ is closed. It follows $X = \{a, b\}$. Contradiction.]
$X$ has a dispersion point.
[Let $m = max\{|A|: A \text{ is an antichain in X} \}$.
By Dilworth's theorem there exists a partition
$X = \bigcup_{i=1}^m C_i$ of $X$ into chains.
By 1., $|C_i| <= 2$ for each i, and, again by biconnectness, there is at most one $C_i$ with two elements.
Hence,
$ n = |X| = \sum _{i=1}^m |C_i| <= 2 + (m-1) = m+1$.
Thus, there is an antichain $A$ of size $n-1$.
It follows, that $A$ is discrete, hence the unique element in $X \setminus A$ is dispersion point.]
Addendum
Thanks to @M W for reminding me to explicitly mention the following fact and a proof thereof:
If $X$ is biconnected, then for all connected subsets
$A, B$ of $X$, each with more than one point, one has
$A \cap B \neq \emptyset$.
For a proof see for instance here ("strongly SG").
Best Answer
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.