Consider $L^2(\mathbb{R})$ as an Hilbert space with inner product $(\cdot ,\cdot)$. Define $$\psi_n(x)=e^{-\frac{x^2}{2}}H_n(x),$$ where $H_n(x)$ is the Hermite Polynomials. How do you show that $\{\psi_n(x)\}$ is a orthogonal basis for $L^2(\mathbb{R})$? (We can make it orthonormal by normalize it).
I showed $(\psi_i,\psi_j)=0$ for all $i\neq j$. To conclude the $\{\psi_n(x)\}$ forms a basis. We need to show that $\text{span}(\psi_n(x))$ is a dense subset of $L^2$. How do we prove that?
As you say, you need to prove that $$ \overline{\text{span}(\psi_n(x))} = L^2(\mathbb{R}). $$ For that, it suffices to prove that if $f\in L^2(\mathbb{R})$ satisfies $$ (f, \psi_n) = 0, $$ for all $n$, then $f = 0$.
Note that the linear span of Hermite polynomials is equal to the linear span of all polynomials (look at the degrees of Hermite polynomials), so it's enough to prove that $$ \int_{\mathbb{R}} f(x) x^n e^{-x^2/2} dx = 0, \; \forall n\geq 0, \Rightarrow f \equiv 0. $$ Now, use that the set of continuous functions with compact support is dense in $L^2(\mathbb{R})$, and by Weierstrass theorem, you can approximate a continuous function on a compact domain uniformly by polynomials. Then, take a sequence of continuous functions with compact support $\{g_n\}$ such that $$ g_n \rightarrow f\cdot e^{-x^2/2} \ \ \ \in L^2(\mathbb R) $$ and, for each $g_n$, a sequence of polynomials $P_{n,m}$ such that $$ P_{n,m} \cdot I_{[supp(g_n)]} \rightarrow g_n \quad \text{(uniformly in $m$)}, $$ where $I_{[supp(g_n)]}$ is the indicatrix of the support of $g_n$. Now consider $$ A_{n,m} = \int \left(P_{n,m}(x) \cdot I_{[supp(g_n)]}(x) - f(x)e^{-x^2/2}\right) \cdot f(x) e^{-x^2/2} dx.$$ By the triangular inequality and choosing $m(n)$ large enough for each $n$, we have $$A_{n,m(n)} \rightarrow 0, \quad \text{as }n\to \infty,$$ but also $$ A_{n,m(n)} \rightarrow -\int f(x)^2 e^{-x^2} dx, \quad \text{as }n \to \infty. $$ Then $$ 0 = -\int f(x)^2 e^{-x^2} dx \Rightarrow f \equiv 0 $$ which concludes the argument.
