Prove that a set of elements in a vector space are linearly dependent

Note that you only need to prove that $c_1 + c_2$ and $2c_1 - c_2$ are not both zero to show that $\{X_1,X_2,X_3\}$ is linearly dependent.

To see this, suppose that $c_1 + c_2 = 2c_1 - c_2 = 0$. This would imply that $c_1 = c_2 = 0$. Do you see how this contradicts your assumption that $\{ Y_1, Y_2\}$ is linearly dependent?

In this case I think it would be easiest and cleanest to prove the contrapositive, namely, that if $\{X_1,\dotsc, X_3\}$ is a linearly independent set of vectors, then so is $\{Y_1, Y_2\}$. To this end suppose that $$ c_1Y_1+c_2Y_2=0\tag{1} $$ for some $c_i\in\mathbb{R}$. Since $Y_1=X_1+X_2+2X_3$ and $Y_2=X_1+X_2-X_3$ we can rewrite (1) as $$ (c_1+c_2)X_1+(c_1+c_2)X_2+(2c_1-c_2)X_3=0 $$ whence $c_1+c_2=0$ and $2c_1=c_2$ by linear independence of the $X_i$. In particular $3c_1=0$ whence $c_1=0$ and so $c_2=0$. It follows that the $Y_i$ are linearly independent.

You're on the right track: you have a linear combination $$ d_1 X_1 + d_2 X_2 + d_3 X_3 = 0 $$ where $d_1 = d_2 = c_1 + c_2$ and $d_3 = 2c_1 - c_2$.

What you need to show is that not all of $d_1,d_2,d_3$ are $0$. So, suppose $d_1 = d_2 = 0$. This means $c_1 = -c_2$. In this case, what can you say about $d_3$? (If you can conclude it is non-zero, you are done).