TL;DR: You forgot to normalize the eigenvectors!
Your description of the spectral theorem is incomplete; it requires you to use an orthonormal basis i.e. the basis $\beta=\{|\psi_j\rangle\}$ must be orthonormal, as I previously discussed here. When normalized, the eigenvectors of the $X$ gate form the basis $\left\{{\frac{|0\rangle+|1\rangle}{\sqrt{2}}},{\frac{|0\rangle-|1\rangle}{\sqrt{2}}}\right\}$. Now let's evaluate the terms $\color{red}{\left(\frac{\langle 0|+\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)}$ and $\color{blue}{\left(\frac{\langle 0|-\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)}$. These are actually the eigenvalues of $X$ corresponding to the two eigenvectors, because if $|\psi\rangle$ is a normalized eigenvector (i.e. $\langle \psi|\psi\rangle = 1$ ) of $X$, and $\lambda$ is the corresponding eigenvalue, then $$X|\psi\rangle = \lambda |\psi\rangle \implies \langle \psi|X|\psi\rangle = \langle \psi|\lambda|\psi\rangle= \lambda\langle \psi|\psi\rangle =\lambda.1=\lambda.$$
Now you know that $X$ is the "bit-flip" gate, so it performs $|0\rangle \mapsto |1\rangle$ and $|1\rangle\mapsto|0\rangle$.
$$\therefore \color{red}{\left(\frac{\langle 0|+\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)} = \left(\frac{\langle 0|+\langle 1|}{\sqrt{2}}\right)\left(\frac{|1\rangle+|0\rangle}{\sqrt{2}}\right)$$
$$=\frac{1}{2}(\langle 0|1\rangle + \langle0|0\rangle+\langle 1|1\rangle + \langle 1|0\rangle)$$
$$=\frac{1}{2}(0+1+1+0)=\color{red}{1} \ (\because \text{$|0\rangle$ and $|1\rangle$ are orthonormal})$$
$$\therefore \color{blue}{\left(\frac{\langle 0|-\langle 1|}{\sqrt{2}}\right)X\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)} = \left(\frac{\langle 0|-\langle 1|}{\sqrt{2}}\right)\left(\frac{|1\rangle-|0\rangle}{\sqrt{2}}\right)$$
$$=\frac{1}{2}(\langle 0|1\rangle - \langle0|0\rangle-\langle 1|1\rangle + \langle 1|0\rangle)$$
$$=\frac{1}{2}(0-1-1+0)=\color{blue}{-1} \ (\because \text{$|0\rangle$ and $|1\rangle$ are orthonormal})$$
So now according the spectral decomposition theorem you can represent the $X$ gate as:
$$\boxed{\color{blue}{-1}\left(\frac{|0\rangle-|1\rangle}{\sqrt{2}}\right)\left(\frac{\langle 0|-\langle1|}{\sqrt{2}}\right) + \color{red}{1}\left(\frac{|0\rangle+|1\rangle}{\sqrt{2}}\right)\left(\frac{\langle 0|+\langle1|}{\sqrt{2}}\right)}\tag{*}\label{*}$$
$$\require{cancel}=-\frac{1}{2}\left(\cancel{|0\rangle\langle0|}-|0\rangle\langle1|-|1\rangle\langle0|+\cancel{|1\rangle\langle1|}\right)+\frac{1}{2}\left(\cancel{|0\rangle\langle0|}+|0\rangle\langle1|+|1\rangle\langle0|+\cancel{|1\rangle\langle1|}\right)$$
$$=\color{green}{|1\rangle \langle0| + |0\rangle \langle1|}$$
To convince you that this result is correct let's apply it on an arbitrary qubit state $|\Psi\rangle=\alpha|0\rangle+\beta|1\rangle$:
$$\color{green}{(|1\rangle \langle0| + |0\rangle \langle1|)}(\alpha|0\rangle+\beta|1\rangle)$$
$$=\alpha|1\rangle\langle0|0\rangle+\beta|0\rangle\langle 1|1\rangle$$
$$=\alpha|1\rangle + \beta|0\rangle$$
So yes, the bits $|0\rangle$ and $|1\rangle$ of $|\Psi\rangle$ are flipped and our representation $\eqref{*}$ of the $X$ gate is indeed correct!