I wanted to show that arc-connectedness is an equivalence relation. Checking reflexivity and symmetry is easy. This is my try at proving transitivity:
Let $x\neq y$, $y\neq z$ and $x\neq z$ be three points in $X$ and let $f:[0,1]\to X$ and $g:[0,1]\to X$ be two arcs(i.e., injective continuous functions) such that $f(0)=x$, $f(1)=g(0)=y$ and $g(1)=z$.
Let $S:=f([0,1])\bigcap g([0,1])$. Then $y\in S$ and thus $S \neq \emptyset$. Let $T:=f^{-1}(S)$. Then $1\in T$ and thus $T\neq \emptyset$. Let $c:=$inf $T$, which exists as $T$ is non-empty and bounded below by $0$.
Now as $c=$inf $T$, $\exists \{x_{n}\}_{n\geq 1}\subseteq T$ such that $x_{n}\rightarrow c$ as $n\rightarrow \infty$. But as $f$ is continuous, this means that $f(x_{n})\rightarrow f(c)$ as $n\rightarrow \infty$. Now, as $\forall n\geq 1$, $f(x_{n})\in S$, we have $\forall n\geq 1$, $\exists y_{n}\in[0,1]$ such that $f(x_{n})=g(y_{n})$. Thus $g(y_{n})\rightarrow f(c)$ as $n\rightarrow \infty$. Thus $f(c)\in \overline{g([0,1])}$.
This is where I am stuck in the non-Hausdorff case. In the Hausdorff case, we observe that, as $[0,1]$ is compact and $g$ is continuous and thus, $g([0,1])$ is a compact subspace of the Hausdorff space $X$ and thus, is closed. Thus $f(c)\in \overline{g([0,1])}=g([0,1])$. Thus $f(c)=g(d)$, say for some $d\in [0,1]$. Then, using the "gluing lemma", we observe that the function $h:[0,c+1-d]\to X$ given by,
$$h(\alpha) =
\begin{cases}
f(\alpha) & \text{if $\alpha \leq c$} \\
g(\alpha -c+d) & \text{if $\alpha \geq c$}
\end{cases}$$
is a continuous function with $h(0)=x$ and $h(c+1-d)=z$. Further, our choice of $c$ shows that $h$ is injective. As $[0,1]$ and $[0,c+1-d]$ are homeomorphic, we are done(in the Hausdorff case).
Any help in the non-Hausdorff case will be helpful.
The relation is not transitive in general. Consider the closed interval $[-1, 1]$ with additional copy of $0$ denoted by $0'$ whose neighborhoods have base $\{0'\} ∪ (0, t)$ for $t > 0$. It is the same thing as considering $[-1, 1]$ and another copy of $[0, 1]$ and identifying the corresponding copies of points from $(0, 1]$.
You may prove that this space is a counterexample (or I may write down the proof if you want).