Can this be done using the inner product?
I can successfully do this by applying the Gram-Schmidt process and orthonormalizing the bases, but what if I wanted to project the function onto my non-orthogonal bases?
** I am self-learning, forgive me (and feel free to correct me) if my terminology/understanding seems wrong
If you're trying to project onto a finite-dimensional subspace $\mathcal{M}$ spanned by a basis $\{ v_1,v_2,\cdots,v_n \}$, then you can write down a matrix equation and solve. The projection $P_{\mathcal{M}}x$ of $x$ onto $\mathcal{M}$ has the form $$ P_{\mathcal{M}}x = \alpha_1 v_1 + \alpha_2 v_2 + \cdots +\alpha_n v_n, $$ where the $\alpha_j$ are determined by $$ (x-\alpha_1 v_1 -\alpha_2 v_2-\cdots-\alpha_n v_n) \perp\mathcal{M}. $$ Equivalently, the $\alpha_j$ are determined by the $n$ equations $$ (x-\alpha_1 v_1-\alpha_2 v_2-\cdots-\alpha_n v_n,v_k) =0,\;\; 1 \le k \le n. \tag{$\dagger$} $$ The coefficient matrix is a covariance matrix: $$ \left[\begin{array}{cccc} (v_1,v_1) & (v_2,v_1) & \cdots & (v_n,v_1) \\ (v_1,v_2) & (v_2,v_2) & \cdots & (v_n,v_2) \\ \vdots & \vdots & \ddots & \vdots \\ (v_1,v_n) & (v_2,v_n) & \cdots & (v_n,v_n) \end{array}\right] \left[\begin{array}{c}\alpha_1 \\ \alpha_2 \\ \vdots \\ \alpha_n\end{array}\right] = \left[\begin{array}{c}(x,v_1)\\(x,v_2)\\ \vdots\\ (x,v_n)\end{array}\right] $$ To show that there is a unique solution vector $\alpha$ for every $x$, it is enough to show that the matrix has a trivial null space. But $\alpha$ is a null solution iff $(\dagger)$ holds with $x=0$ for this $\alpha$, which implies that $\alpha_1 v_1 + \alpha_2 v_2 + \cdots \alpha_n v_n = 0$ and, hence, $\alpha_1=\alpha_2=\cdots=\alpha_n=0$ because the $v_j$ are independent. So the covariance matrix has a trivial null space, which implies the existence and uniqueness of a solution $\alpha$ for all $x$.
For an infinite-dimensional subspace $\mathcal{M}$, the strategy must be different because you end up with an infinite set of equations. Given any $x$, a vector $m\in\mathcal{M}$ satisfies $(x-m)\perp \mathcal{M}$ iff $\|x-m\| \le \|x-m'\|$ for all $m'\in\mathcal{M}$, and such an $m$ is unique if it exists. That is, orthogonal projection and closest-point projection are the same; one exists iff the other does, and the two are the same if either exists. This is true regardless of dimension. One can always choose $\{ m_n \}$ is chosen so that $$ \|x-m_n\| < \inf_{m\in M}\|x-m\|+\frac{1}{n}, $$ and it turns out that $\{ m_n \}$ is a Cauchy sequence. This sequence converges to some $m$ iff there exists such a projection as described and, in this case, the projection is $m$. Therefore, a projection $P_{\mathcal{M}}$ exists if $\mathcal{M}$ is complete (i.e., Cauchy sequences in $\mathcal{M}$ converge to something in $\mathcal{M}$.) That happens, for example, if $\mathcal{M}$ is a closed subspace of a Hilbert space. Or it happens if $\mathcal{M}$ is a finite-dimensional subspace, as shown above.