Bounding the norm of an extension of a linear function from a subspace of a normed space to a finite dimensional normed space

I got this problem in my homework. Let $X$ be a normed space, and let $X_0\subset X$ be a subspace.

Let $T$ be a continuous linear function from $X_0$ to $Y$, where $Y$ is an $n$-dimensional normed space.

Prove that $T$ can be extended to a continuous linear function $T'$ on $X$ s.t. $\|T'\|\le n\|T\|$

In my attempt to solve this problem, I took an Auberbach basis $\{e_i\},\{e^i\}$ of $Y$. I assumed WLoG that $T$ is onto $Y$ and for each $e_i$ chose an $x_i\in X_0$ s.t. $Tx_i=e_i$.

I then examined $\operatorname{Span}\{x_i\}$, this is an $n$-dimensional subspace of $X$, and I can thus define a projection $P$ from $X$ to $\operatorname{Span}\{x_i\}$ such that $\|P\|\le n$. I get that the composition $T\circ P$ is bound as needed, satisfies all the conditions and identifies with $T$ on $\operatorname{Span}\{x_i\}$. However, I wasn't able to prove that implies that it identifies with $T$ on the entire $X_0$. Furthermore, I wasn't even able to convince myself that it's true.

Any suggestions?

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Do you mean, $T:X_0 \to Y$ instead of $T:X \to Y$? –  math Jan 7 '12 at 9:35
affirmative.... –  Shai Deshe Jan 7 '12 at 9:37
Hint: Reduce to the case $Y = \mathbb{R}^n$ with the $\|\cdot\|_\infty$-norm for which Hahn-Banach allows you to extend norm-preservingly. –  t.b. Jan 7 '12 at 9:45
But... Hahn-Banach only works for functionals (I know I accidentally wrote that T is a functional but it clearly isn't). –  Shai Deshe Jan 7 '12 at 9:47
Choose a basis $e_1,\ldots,e_n$ for $\mathbb{R}^n$ and observe that $T = \sum T_k e_k$ for some linear functionals $T_k:X_0 \to \mathbb{R}$. Extend those... –  t.b. Jan 7 '12 at 9:48

Choose a basis $e_1,\ldots,e_n$ of unit vectors of $Y$ with dual basis $e^1,\ldots,e^n$. Put $T_k = e^k \circ T: X_0 \to \mathbb{R}$. Observe that $\|T_k\| \leq \|T\|$. We can extend $T_k$ to a functional $S_k: X \to \mathbb{R}$ with $\|S_k\| = \|T_k\|$ by Hahn-Banach. Now put $S = \sum_{k=1}^n S_{k} \cdot e_k : X \to Y$. To see that $S$ extends $T$ just note that for $y \in Y$ we have $y = \sum e^k(y) \cdot e_k$ so that $Tx_0 = \sum e^k(Tx_0) \cdot e_k = \sum S_k(x_0)\cdot e_k$ for all $x_0 \in X_0$.
Finally, $\|S\| \leq \sum \|S_k\| \leq n \|T\|$.