The DTFT of $\{1,1\}$ is $1+e^{-j\omega}$ but what is the DTFT of $\{1,-1\}$?
You should consider the DTFT pair: 


  $$\delta[n] \xrightarrow{\text{DTFT}} 1$$


and the time-shifting property of the Fourier transform


  $$x[n-n_0] \xrightarrow{\text{DTFT}} X(e^{j\omega})e^{-j\omega n_0}$$


plus the linearity of the FT. That is,


  $$ax_1[n]+bx_2[n] \xrightarrow{\text{DTFT}} aX_1(e^{j\omega})+bX_2(e^{j\omega})$$




Now consider the discrete signal $\{1,1\}$ which can be represented by $x[n], n\in\mathbb{Z}$ as follows 
$$x[0]=1\\x[1]=1\\x[n]=0, \forall n\in\mathbb{Z} \ \backslash \ \{0,1\}$$

and can be denoted by $x[n]=\delta[n]+\delta[n-1]=x_1[n]+x_2[n]$. The DTFT is $$1+(1)e^{-j\omega(1)}=1+e^{-j\omega}$$
Similarly, the signal $\{1,-1\}$ can be represented by 
$$x[0]=1\\x[1]=-1\\x[n]=0, \forall n\in\mathbb{Z} \ \backslash \ \{0,1\}$$

which can be denoted by $x[n]=\delta[n]-\delta[n-1]=x_1[n]-x_2[n]$, and the DTFT is $$1-(1)e^{-j\omega(1)}=1-e^{-j\omega}$$