Consider the subspace $Y =\{(x, x): x\in \mathbb{C}\}$ of the normed linear space $(\mathbb{C}^2, \|\ \|_{\infty})$. If $\phi$ is a bounded linear functional on $Y$, defined by $\phi(x, x)=x$, then which one of the following sets is equal to $$\big\{\psi(1, 0): \psi \ \text{is a norm preserving extension of}\ \phi\ \text{to}\ (\mathbb{C}^2, \|\ \|_{\infty})\big\}, \ \|(x, y)\|_{\infty}= \sup\{|x|, |y|\} $$
$\{1\}$
$[0, 1]$
My attempt: It can be easily shown that $\|\phi\|_{\infty}=1$. Since $\psi$ is a norm preserving extension, so $$\|\psi\|_{\infty}= \|\phi\|_{\infty}=1$$ Now as $\|(1, 0)\|_{\infty}= 1$, so this means $\psi(1, 0)\leq 1$. How to conclude further?
Any linear functional on $\mathbb C^2$ is of the form $\psi((x,y)) = a x + b y$ for some constants $a$ and $b$. This is easily seen to have norm $|a|+|b|$. Now you want $ax + b x = x$, i.e. $a+b=1$, and $|a|+|b|=1$, and you want to know what can $a$ be. If $a \in [0,1]$, what should you take for $b$?