If a topological subspace $Y$ of $X$ is simply connected (but $X$ isn't) and we extended $Y$ with a path, meaning we take the union of $Y$ and a path from a point in $Y$ to a point in $X$ outside $Y$ (in a straight line, or at least not in a loop), is the extension of $Y$ then also simply connected?
I think so because the extension is still path connected and all loops with portions of the path in it is still be homotopic to the trivial loop?
but how to prove this in formal way?
The answer is no. Consider the unit open ball around the origin in $\mathbb R^2$, which is clearly simply connected. Extend a path from $(0,0)\to(1,0)$ that goes to $(0,1)$, then to $(1,1)$, then to $(1,0)$ in straight segments. This new domain isn't simply connected, as the square with corners $(0,0)$, $(1,0)$, $(1,1)$, and $(0,1)$ is a closed curve that can't be reduced to a point.
Edit: OP wants an example with a straight path, the picture below illustrates an example:
Another example: Let $B(x,r)$ be the open ball of radius $r$ around $x$ in $\mathbb R^2$. Consider $C=B((0,0),2)\setminus\overline{B((0,0),1)}$ and the path $(1.5,0)\to(-1,0)$