Show that if the space $\bar{A}-$Int $A$ is path-connected, then the space $\bar{A}$ is path-connected. 
Let $A$ be a subset of a path-connected topological space $X$. Show that if the space 
  $\bar{A}-$Int $A$ is path-connected, then the space $\bar{A}$ is path-connected.

My Try::
Let $a,b\in \bar{A}$. It suffices to consider only the cases where, 
1) $a,b\in $ Int $A$ and 
2) $a\in $Int $A,b\in \bar{A}-$Int $A$
Regarding case 1), since $X$ is path connected there is a path $\alpha:[0,1]\rightarrow X$ such that $\alpha(0)=a$ and $\alpha(1)=b$. After that I could give a pictorial proof as given in the picture below, but it was very difficult me to convert it to words. Can anybody please help me to write a complete proof in words?

 A: If the image of path $\alpha$ is contained in $\bar A$ then we are done. If not then consider $s=\inf\{t: \alpha(t) \notin \hbox{Int } A\}$, $x=\alpha(s)$, $t=\sup\{t: \alpha(t) \notin \hbox{Int } A\}$, $y=\alpha(t)$. Note that $x,y \in \bar A - \hbox{Int } A$. 
Since $\bar A - \hbox{Int } A$ is path-connected, there exists a path $\beta:[0,1] \to \bar A - \hbox{Int } A$ such that $\beta(0)=x$ and $\beta(1)=y$. 
Now define path $\gamma:[0,1] \to X$ by the following formula:
$$\gamma(u)=\begin{cases} 
\alpha(3us) \qquad\qquad\qquad 0\le u<\frac 13 \\ 
\beta(3u-1)  \qquad\qquad\qquad \frac 13 \le u < \frac 23 \\
\alpha(3u-2+3(1-u)t) \qquad \frac 23 \le u \le 1
\end{cases}$$
It is straightforward to check that its image lies in $\bar A$ and $\gamma(0)=a, \gamma(1)=b$.
A: If $\text{Im }\alpha$ is not already contained in $\bar{A}$ then some part of it is in $X-\bar{A}$. By the continuity of $\alpha$ the image should intersect the boundary of $\bar{A}$ which is by definition $\bar{A} \cap \overline{X-\bar{A}}$ and is clearly contained in $\bar{A} - \text{Int} A$.
Let us consider two points $x_0$ and $x_1$ in (boundary of $\bar{A}$) $\cap$ $ \text{Im  }\alpha$. (note that $x_0$ can be equal to $x_1$). Then since these points are in $\bar{A}-\text{Int}A$ there is a path $x_0x_1$ in $\bar{A}-\text{Int}A$. 
Let $ax_0$ be the part of $\alpha$ from $a$ to $x_0$ and $x_1b$ be the part of $\alpha$ from $x_1$ to $b$. Then the path in $\bar{A}$ from $a$ to $b$ is $ax_0*x_0x_1*x_1b$ the concatenation of the three paths.
You can do case 2 in a similar way.
I hope this was rigorous enough for you. 
