The answer is no.
Let be construct such a $(V,\sigma)$. Let $X$ be the set of functions $\mathbf{N}\to\mathbf{R}$. Let $Y\subset X$ be a subset such that
- $Y$ is linearly independent
- $Y$ contains all Dirac functions $n\mapsto\delta_{m,n}$
- for every $g\in X$, there exists $f\in Y$ such that $\limsup(f/(|g|+1))=\infty$.
Note that the last condition forces $Y$ to be uncountable. The existence of $Y$ is checked by considering a maximal linearly independent subset containing all Dirac functions.
Now consider the space $V$ with basis indexed by $\mathbf{N}\sqcup Y$, namely $(e_n)_{n\in\mathbf{N}}$ and $(z_f)_{f\in Y}$, and symmetric bilinear form $$\sigma(e_n,e_m)=\delta_{m,n},\quad \sigma(e_n,z_f)=f(n),\quad\sigma(z_f,z_g)=0.$$
I claim that $\sigma$ is non-degenerate; I'll use the first two axioms. Indeed, let $v=\sum_na_ne_n+\sum_fb_fz_f$ belong to the kernel of $\sigma$. Then $0=\sigma(v,z_f)=\sum_na_nf(n)=0$ for all $f\in Y$. Choosing $f=\delta_n$, we deduce $a_n=0$, for all $n$. So $v=\sum_fb_fz_f$. Then $0=\sigma(v,e_n)=\sum_{f\in Y}b_ff(n)$ for all $n$, which means that $\sum_{f\in Y}b_ff=0$. Since $Y$ is linearly independent, we deduce $b_f=0$ for all $f$, and hence $v=0$.
Now suppose the existence of $H$, $S:V\to H$ and $T$ as required. Write $E_n=S(e_n)$ and $Z_f=S(z_f)$, $C_f=\|TZ_f\|$, $g(n)=\|E_n\|$. Then for all $f\in Y$ and $n$, we have $f(n)=\sigma(z_f,e_n)=\langle TZ_f,E_n\rangle$. So $|f(n)|\le C_fg(n)$ for all $n$. Hence $\limsup (f/(g+1))<\infty$. Since this holds for each $f\in Y$, we contradict the third axiom, which ends the proof.
However, if $V$ has countable dimension, the answer is yes. Indeed, one can construct, by induction, an orthogonal basis and conclude in the same way as in finite dimension.
For the induction, one first observes that every finite-dimensional subspace $W$ is contained in a non-degenerate one. This allows to write $V$ as an increasing union of non-degenerate subspaces, and thereby construct the desired basis.
