Hahn-Banach extension uniqueness. Given $f:C[0, 1]\to\mathbb{R}$ - a linear function, defined on 1d linear span with $x(t)=t+1$ as basis.
Find out if there is unique Hahn-Banach extension preserving norm on the whole $C[0, 1]$. 
I have tried to recall to Riesz representation theorem, so that there exists unique $y_1$ in span and for all $x$ from the span $f(x)=\langle x, y_1\rangle
$ and $\|f\|=\|y_1\|$. Same goes for the extension $F$ where the unique element is $y=y_1+y_2$ where $y_2\in\langle t+1\rangle^\bot$, $\|F\|=\|y\|=\|f\|=\|y_1\|$.
Next steps are unclear.
Thank you in forward!
 A: You seem to be working with the $\|\cdot\|_2$ norm.
You're on the right track. Since $Y = \operatorname{span}\{t \mapsto t+1\}$ is a finite-dimensional inner product space, it is a Hilbert space. Hence a functional $f : Y \to \Bbb R$ can be written as $f(x) = \langle x, y_1\rangle$ for some $y_1 \in Y$, by the Riesz representation theorem.
We claim that $F : C[0,1] \to \Bbb R$ given by $F(x) = \langle x, y_1\rangle$ is the unique Hahn-Banach extension of $f$.
Namely, clearly $F$ extends $f$ and $\|F\| = \|y_1\|_2 = \|f\|$. Assume that $G : C[0,1] \to \Bbb R$ is also a Hahn-Banach extension of $f$. $C[0,1]$ is not a Hilbert space but its completion $L^2[0,1]$ is a Hilbert space, so there exists $z \in L^2[0,1]$ such that $G(x) = \langle x, z\rangle$ for all $x \in C[0,1]$.
Let $z = z_1 + z_2$ where $z_1 \in Y, z_2 \in Y^\perp$.
Now for all $y \in Y$ we have
$$\langle y, y_1\rangle = f(y) = G(y) = \langle y, z\rangle = \langle y, z_1\rangle \implies y_1 = z_1$$
because $G$ extends $f$.
On the other hand
$$\|y_1\|_2^2 = \|f\|^2 = \|G\|^2 = \|z\|_2^2 = \|z_1\|_2^2 + \|z_2\|_2^2 = \|y_1\|_2^2 + \|z_2\|_2^2$$
hence $z_2 = 0$ so we conclude $z = y$ and hence $G = F$.
