Bounded projection onto finite dimensional subspace of normed space

Let $X$ be a normed linear space and $X_0$ be a finite dimensional subspace of $X$. Prove that there exists a projection operator $P \in B(X)$ such that $R(P ) = X_0$.

My Try: $X$ can always be written as $X_1 + X_0$ where $X_1$ is a closed subspace and $X_0 \cap X_1 = \{0\}$. So $x \in X, x = y+z, y \in X_0, z\in X_1.$ Define the map $P(x)=y$. I think that this map works.

Can someone check the argument is correct or not? And how to show $P \in B(X)$?

Let $X_0 \le X$ be a finite-dimensional subspace of $X$ and let $\{e_1, \ldots, e_n\}$ be a basis for $X_0$. Then every $x \in X_0$ has a unique representation as a linear combination of $e_1, \ldots, e_n$:

$$x = \alpha_1(x)e_1 + \ldots \alpha_n(x)e_n$$

Notice that $\alpha_i : X_0 \to \mathbb{F}$ are linear functionals, and they are bounded because since functionals on a finite-dimensional space are always bounded.

Thus, using the Hahn-Banach theorem, we can extend them to bounded functionals $\tilde\alpha_i : X \to \mathbb{F}$.

Now define $P : X \to X_0$ as

$$Px = \tilde\alpha_1(x)e_1 + \ldots \tilde\alpha_n(x)e_n, \quad x\in X$$

Notice that $P$ acts as identity on $X_0$:

$$Px = \tilde\alpha_1(\underbrace{x}_{\in X_0})e_1 + \ldots \tilde\alpha_n(\underbrace{x}_{\in X_0})e_n = \alpha_1(x)e_1 + \ldots \alpha_n(x)e_n = x, \quad x \in X_0$$

Therefore, \DeclareMathOperator{\Ima}{Im}P^2 = P since \Ima P = X_0. Finally, P is bounded: \begin{align} \|Px\| &= \|\tilde\alpha_1(x)e_1 + \ldots \tilde\alpha_n(x)e_n\| \\ &\le \|\tilde\alpha_1(x)\|\|e_1\| + \ldots \|\tilde\alpha_n(x)\|\|e_n\| \\ &\le \|\tilde\alpha_1\|\|x\|\|e_1\| + \ldots \|\tilde\alpha_n\|\|x\|\|e_n\| \\ &= \big(\|\tilde\alpha_1\|\|e_1\| + \ldots \|\tilde\alpha_n\|\|e_n\|\big)\|x\| \end{align} Thus, P is the desired projection. From \DeclareMathOperator{\Ker}{Ker}here we also get the decomposition X = X_0 \dot+ \Ker P = X_0 \dot+ \bigcap_{i=1}^n \Ker\tilde\alpha_i. Thus follows that a finite-dimensional subspace always has a direct complement, a fact which cannot be used before conducting a proof similar to this one. It's not true that X can always be decomposed in this way for any given closed subspace. It's critical that X_0 is finite dimensional. Since X_0 is finite dimensional, it is isomorphic to a Hilbert space of finite dimension, say n. If e_1,\ldots,e_n is an orthonormal basis for X_0 with the inner product \langle \cdot,\cdot\rangle, then the mapping x\mapsto \langle x,e_i\rangle is a bounded linear functional on X_0 for each i=1,\ldots,n. Using the Hahn-Banach theorem we can extend each of these to a bounded linear functional \phi_i on X. Now define P:X\to Y by P(x)=\sum_{i=1}^n\phi_i(x)e_i. Since every x\in X_0 satisfies x=\sum_{i=1}^n\langle x,e_i\rangle e_i=P(x), we infer that P is a projection onto X_0. Moreover, P is bounded because \|Px\| \le\sum_{i=1}^n\|\phi_i(x)\|\le\sum_{i=1}^n\|\phi_i\|\|x\|.\$