Yes, I think this is true. Any infinite dimensional Banach space $V$ contains a basic sequence $(x_n)$. Then $\{x_1, x_3, x_5, \ldots\}$ is linearly independent and therefore its closed span $V_0$ is infinite dimensional. The codimension of $V_0$ must be infinite, otherwise there would be a linear dependence among $\pi(x_2), \pi(x_4), \pi(x_6), \ldots$ where $\pi: V \to V/V_0$ is the natural projection, and this would lift to make some finite linear combination of $x_2, x_4, x_6, \ldots$ with at least one nonzero coefficient belong to $V_0$, which is impossible.
New edit: Bruce has pointed out that it isn't obvious that a nonzero finite linear combination of $x_2, x_4, x_6, \ldots$ cannot belong to $V_0$. Indeed it isn't obvious, but in fact this cannot happen. Let $x = a_2x_2 + \cdots + a_{2n}x_{2n}$ be a finite linear combination with at least one nonzero coefficient. Then its distance to ${\rm span}(x_1, \ldots, x_{2n-1})$ is strictly positive, greater than some $\epsilon > 0$. Thus if we fill out $a_2, \ldots, a_{2n}$ to a sequence $(a_n)$ in any way, the norm of $a_1x_1 + a_2x_2 + \cdots + a_{2n}x_{2n}$ must be at least $\epsilon$.
Now we use the fact that since $(x_n)$ is a basic sequence there exists $K > 0$ such that $\left\|\sum_{i=1}^{2n} a_ix_i\right\| \leq K\left\|\sum_{i=1}^m a_ix_i\right\|$ for any $m \geq 2n$ (see the first answer to this question). This shows that the span of any finite subset of $\{x_1, x_3, x_5, \ldots\}$ is at least $\epsilon/K$ away from $x$, and therefore $x$ cannot belong to $V_0$.
Best Answer
For $n :=\{0,1,\dots,n-1\}\in\omega$ let’s denote $P_n:\mathbb{R}^\omega\to \mathbb{R} ^n$ the projection, which is the restriction map $f\mapsto f_{|n}$. The dimensions of the subspaces $P_n(L)\subset \mathbb{R} ^n$ can’t be stationary, because that would mean that for some $n_0$, every function $f\in P_{n_0}(L)$ has a unique extension to a function $\tilde f\in L$, in which case $ P_{n_0}(L)\ni f\mapsto \tilde f\in L$ is a linear bijection, whereas $L$ was assumed infinite-dimensional. So let $(n_k)_k$ be a strictly increasing sequence such that $\dim P_{n_k+1}(L)> \dim P_{n_k}(L)$. Thus $\ker \big(P_{n_k}: P_{n_k+1}(L)\to P_{n_k}(L) \big)\neq(0)$: there is a function $f_k\in L$ such that $f_k|{n_k}=0$ and $f_k(n_k+1)=1$. So if we take $ I:=\{n_k+1\}_{k\in\omega}$, for any infinite $J\subset I$ The space $P_J(L)$ contains the functions $P_Jf_k={f_k}_{|J}$, which are clearly linearly independent.