Convex subset

Your proof sounds ok but you seem to be proving that $A_1\cap A_2$ is convex directly, and not using the fact that the intersection of two convex set is convex. Below is how I would prove it. 


Let $A_1$ and $A_2$ be defined as in your notes.  Since 
\[A=A_1\cap A_2,\]
to show that $A$ is convex, it is enough to show that both $A_1$ and $A_2$ are convex. 


Let $(x_1,y_1), (x_2,y_2)\in A_1$. Then 
\[(1)       y_1 \leq x_1   \text{and}     \ \ y_2 \leq x_2. \]
Thus from (1)  we get 
\[ty_1+(1-t)y_2 \leq tx_1 +(1-t)x_2 \]
which implies that for $t\in [0.1]$
\[t(x_1,y_1)+(1-t)(x_2,y_2) = (tx_1 +(1-t)x_2, ty_1+(1-t)y_2) \in A_1.\]
Hence $A_1$ is convex. 

Next we show that $A_2$ is also convex. Let $(x_1,y_1), (x_2,y_2)\in A_2$. Then
\[(2)      y_1 \leq 2    \text{and}   \ \ y_2 \leq 2. \] It follows from (2) that \[ty_1+(1-t)y_2 \leq 2t+2(1-t)=2\]
which implies that for $t\in [0.1]$ \[t(x_1,y_1)+(1-t)(x_2,y_2) = (tx_1 +(1-t)x_2, ty_1+(1-t)y_2) \in A_2.\] Hence $A_2$ is convex. 

Since both $A_1$ and $A_2$ are convex, the intersection $A=A_1\cap A_2$ is also convex.