Let $U$, $V$ be two open disjoint sets over $X$ such that $X=U\cup V$. Let $p\in U$. For any $q\in V$, let $\alpha :[0,1]\rightarrow X$ be a path connecting $p$ and $q$. Then the set $A:=\alpha^{-1}(U)$ is open in $[0,1]$. Let $x:=\sup A$. Then $y=\alpha (x)\in \partial U$:
- If $W$ is an open set with $\alpha(x)\in W$, then $\alpha^{-1}(W)\cap A\neq \emptyset \implies \alpha(\alpha^{-1}(W)\cap A)=W\cap \alpha(A)=W\cap U\neq \emptyset$.
- Likewise, $\alpha^{-1}(W)\cap ([0,1]-A)\neq \emptyset\implies \alpha(\alpha^{-1}(W)\cap ([0,1]-A))=W\cap (\alpha([0,1])-U)\subset W\cap V\neq \emptyset$.
Now, if $\alpha(x)\in V$, then $U\cap V=\emptyset \implies \alpha(x)\notin \partial U$ (which is a contradiction). Then $\alpha(x)\in U$. For every $y\in \partial U$, we can find a path (through path concatenation) such that $\alpha (x)=y$ for some $x\in [0,1]$, so $y\in U \implies U=\overline U$. That is, $U$ is open and closed in $X$. Then $\alpha^{-1}(U)$ must be open and closed in $[0,1]\implies \alpha^{-1}(U)=[0,1]$ (because $[0,1]$ is connected). Thus $q\in U$ for every $q\in V$, which is a contradiction, and $V$ must be empty. Therefore, $X$ is connected.
You know that $[0,1]$ is connected, so the image of $\alpha$ is connected. On the other hand, $$ \alpha([0,1])=\bigl(\alpha([0,1])\cap U\bigr)\cup\bigl(\alpha([0,1])\cap V\bigr) $$ is written as the disjoint union of two relative open sets. Connectedness forces either $\alpha([0,1])\cap U=\emptyset$ or $\alpha([0,1])\cap V=\emptyset$. This is a contradiction, because $\alpha(0)=p\in U$ and $\alpha(1)=q\in V$.