CLAIM:(note the edit at the bottom)
Let $X$ be a normed space, $\{x_1,x_2, \cdots \}\subseteq X$ a linearly independet subset and $\alpha_1,\alpha_2,\cdots \in\mathbb{R}$. Then there exists $f\in X^*$ such that $f(x_i)=\alpha_i$ for all $i\in\mathbb{N}$.
I know that this is an immediate consequence of Hahn-Banach's theorem. But I only know how to prove the claim when the linearly independent subset is finite, and the obvious generalization of this proof doesn't hold. I will enunciate from which version of HB's theorem I wish to prove the claim...
HB's Theorem:
Let $X$ be a normed space and $S\subseteq X$ a subspace.
Then, given $\varphi\in S^* $ there is $\widetilde{\varphi}\in X^*$ such that $\left.\widetilde{\varphi}\right|_S = \varphi\ $ and $\ \|\widetilde{\varphi}\|=\|\varphi\|$.
Proof of the claim when the linearly independent subset is finite: Just consider $S=[x_1,\cdots,x_n]\subseteq X$ and use HB to extend $\varphi \in S^*$ defined by $\varphi(x_i)=\alpha_i$.
BUT, in the case when the linearly independent subset is $\{x_1,x_2, \cdots \}$ the linear function defined by $\varphi(x_i)=\alpha_i$ is not bounded, so it doesn't belong to $S^*$... and I couldn't prove the claim in another way :(
Could you help me?
Edit: so it happens that the claim is false, as pointed out by @peek-a-boo 's answer. So I'm searching for additional hypothesis on the real sequence $\{\alpha_n\}$. I could assume that it is bounded, or even that the series $\sum \alpha_n$ converges... but I still can't prove that the function which satisfies $\varphi(x_i)=\alpha_i$ is bounded.
Isn't the claim false? Suppose WLOG that $\{x_n\}_{n=1}^{\infty}$ is a collection of linearly independent unit vectors, and $\{\alpha_n\}_{n=1}^{\infty}$ is an unbounded sequence of real numbers (for example $\alpha_n=n$). Then, for any linear $f:X\to\Bbb{R}$ such that for each $n\in\Bbb{N}$, $f(x_n)=\alpha_n$, we have \begin{align} \sup_{x\in X, \|x\|=1}|f(x)|\geq \sup_{n\in\Bbb{N}}|f(x_n)|=\sup_{n\in\Bbb{N}}|\alpha_n|=\infty. \end{align} So, $f$ is unbounded.