Linear transformation $T$ such that for every extension $\overline{T}$, $\|\overline{T}\|>\|T\|$. Let $E$ and $F$ be normed spaces such that $\dim F < \infty$, $G$ a subspace of $E$ and $T:G\rightarrow F$ a continuous linear map. I know that there exists a continuous linear extension $\overline{T}:E\rightarrow F$. Also, if $E$ is a Hilbert space, then $\overline{T}$ can be chosen in the way that $\|\overline{T}\|=\|T\|$.

Problem: Find an example of $E$, $F$, $G$ and $T$ (like above) such that every continuous linear extension $\overline{T}$ has a greater norm, i.e. $\|\overline{T}\|>\|T\|$.


Now, $F$ must be at least a 2-dimensional space, otherwise I could use Hahn-Banach to find an extension with equal norm.  My professor told me it could be done with $E$ of finite dimension. Of course, I tried to come up with an example of $E$ with a norm that doesn't satisfy the parallelogram law. For example, $E=\left(\mathbb{R}^3,\|\cdot\|_{\infty}\right)$ and $F=\left(\mathbb{R}^2,\|\cdot\|_1\right)$. But I couldn't prove that it works with any example I tried using those spaces.
Can somebody help me to find an example and assure that it really has that property?
EDIT:
Apparently, it can't be done with $E$ of finite dimension nor with $F$ equipped with the $\sup$ norm, as @Hamza proved below.
 A: Let $E=\mathbb R^3$, equipped with the sup norm, namely
$$
  \|(x,y,z)\|_\infty  = \max\{|x|, |y|, |z| \},
  $$
and let
$$
  F=G=\{(x, y, z)\in E : x+y+z=0\},
  $$
equipped with the induced norm from $E$.  Also let
$$
  T:G\to F
  $$
be the identity function.  Then clearly $\|T\|=1$, but any extension $\bar T:E\to F$ has norm strictly bigger than
one.
The reason is that any such  $\bar T$ is necessarily  a projection from $E$ to $G$ and there is no
such projection with norm $1$.
The best way to convince oneself of this fact  is to make a cardboard
model of this cube

cut it along the red line, place one of the two halves on top of the table with the red hexagon down,

and attempt to shine a flashlight so that the shadow is restricted to within the hexagon.  It is impossible!

This is based on another answer I recently gave to this question.
A: If $G$ have a topological complimentary in  $E$ (in particular if $E$ is finite dimensional space) i don't think that such construction can be done, because prolonging by $0$ on the complementary space, will have the same norme :
$$
\|\hat{T}\|=\sup_{\begin{array}{c} x\in E \\ x\neq 0\end{array}} \frac{\|\hat{T}x\|}{\|x\|}=\sup_{\begin{array}{c} x=z+y \\ y,z\neq 0\end{array}} \frac{\|\hat{T}z+\hat{T}y\|}{\|z+y\|}=\sup_{\begin{array}{c}  z\in G\\y\in E-G \\ y,z\neq 0\end{array}} \frac{\|Tz\|}{\|z+y\|}=\sup_{\begin{array}{c}  z\neq 0\end{array}} \frac{\|Tz\|}{\|z\|}=\|T\|
$$
If now $E$ is infinite dimensional and $G$ a closed subspace but with no complaiment in $E$, let for example :
$$
\begin{array}{}
I&:& c_0& \to &l_\infty\\
 & & x &\mapsto & x
 \end{array}
$$
so by a famous Phillips's lemma $I$ can't be extended to a continuous linear maps in $l_\infty$.
but if $F$ is finite dimensional space i don't have an explicit example but as remarks you can't take the infinite norm as norm to $F$ in fact, let $T :G \to \mathbb{R}^n $, and let $(T_1,T_2,\dots,T_n)$ it composite then it's clear that $\|Tx\|=\sup_{i\leq n}|T_ix|$, so let $\hat{T}=(\hat{T_1},\dots,\hat{T}_n)$ be the preserving norm extension given by Hahn-Banach theorem   :
$$
\|\hat{T}\|=\sup_{ x\in S_E } \|\hat{T}x\|=\sup_{ x\in S_E } (\sup_{i\leq n}\|\hat{T}_ix\|)=\sup_{i\leq n}(\sup_{ x\in S_E } \|\hat{T}_ix\|)=\sup_{i\leq n} \|T_i\|)=\|T\|
$$ 
 we call this propriety the injectivity of Banach space, and it prove that a finite dimensional space is injective if and only if he is isomorphe isometrically to $l^\infty_n$ 
