Let $A$ be a convex subset of $\mathbb R^n$ and $B$ be a path connected subset of $\mathbb R^n$ such that $\bar A \cap B$ is non empty ; then is it true that $A\cup B$ is path connected ?
Since $\bar A$ is convex , I can see that $\bar A \cup B$ is path connected, but I am not sure about $A\cup B$. I feel that if $x \in \bar A \cap B$ and $y \in A$ then $ty+(1-t)x \in A, \forall t \in (0,1]$ , but I am unable to prove it.
Please help. Thanks in advance.
Your claim is (almost) correct.
You can prove that, if $x\in \overline{A}$ and $y\in \text{relint} A$, then $z_t := ty + (1-t) x\in \text{relint} A$ for every $t\in (0, 1]$.
(For a proof see e.g. Schneider, "Convex bodies...", Lemma 1.1.8.)
The claim is not true if you take merely $y\in A$. Consider, for example, the set in $\mathbb{R}^2$ $$ A := [-1,1]^2 \setminus ((0,1]\times \{1\}). $$ Let $x = (1,1) \in \overline A$ and $y = (0,1) \in A$. The points $tx + (1-t)y$ does not belong to $A$ for $t\in (0,1)$.
Anyway, we can conclude that $A\cup B$ is path connected. Indeed, it is enough to fix a point $y\in\text{relint} A$ so that any point $z\in A$ can be connected to $x$ following the concatenation of the two segments $[z,y]$ and $[y,x]$.