Let $H$ be a infinite dimensional, separable Hilbert space over $\mathbb{C}$
Let $B$ a subset of $H$ such that $B$ is linearly independent and such that exists a homeomorphism $f : [0,1] \to B$ between the unit interval $[0,1]$ and $B$
I would like to know if is it true that:
For every fixed $x \in (0,1)$ $$ \overline{ \operatorname{span} f([0,x]) } \cap \overline{ \operatorname{span} f((x,1]) } = \operatorname{span} f(x) $$
It's not true.
Edit: thinking twice there's a simpler solution.
On $L^2([0,1])$, define $f(x)(t)=1+t^{1/2x}$ for $x\in\mathopen]0,1]$ and $f(0)=1$. It's free (because the $t\mapsto t^x$, $x\ge 0$, form a free family). It depends continuously on $x$ (check at $x=0$). For $x\le 1/2$ it contains all (restrictions of) polynomials hence spans a dense subspace. So for $x=1/2$ the span intersection condition does not hold.
Previous solution (in the lines of my initial comment).
In $L^2([-1,1])$, define, for $0\le t\le 1/2$, $f(t)=\mathbb{1}_{[-1,2t]}$.
It's clearly algebraically free. The closure of its span is $\mathbf{C}\mathbb{1}_{[-1,0]}\oplus L^2([0,1])$ (indeed it contains, for $s<t$, $f(t/2)-f(s/2)=\mathbb{1}_{[s,t]}$, etc).
Let $K$ be a Cantor subset in $[0,1]$ of measure $1/2$. There exists an increasing (hence measurable) map $g:[0,1]\to K$ such that for every $t$, the measure of $[0,g(t)]\cap K$ is equal to $t/2$. Define, for $t\in [1/2,1]$, $$f(t)=\mathbb{1}_{[-1,1]}+\mathbb{1}_{[0,g(2t-1)]\cap K}.$$ Then $f$ is continuous on $[0,1]$.
Clearly it does not satisfy the span intersection condition at $x=1/2$.
What remains is to show that $(f(t))_{0\le t\le 1}$ is free. Consider a nontrivial linear relation: it can be written as $$\sum_{i=1}^na_if(t_i)=\sum_{j=1}^mb_jf(u_j),\quad\text{ with }0\le t_1<\dots<t_n\le 1/2,\; 1/2<u_1<\dots<u_m\le 1,$$ with all $a_i,b_j$ nonzero and $\max(m,n)\ge 1$. Since $(f(t))_{0\le t\le 1/2}$ is free, $m\ge 1$; write the linear dependence assumption as the equality in $L^2([-1,1])$: $$\sum_{i=1}^na_if(t_i)-\sum_{j=1}^{m-1}b_jf(u_j)=b_mf(u_m).$$ The left-hand term is piecewise constant on $[g(u_{m-1}),1]$ (or $[0,1]$ if $m=1$) while the right-hand term is not: if it were, $\mathbb{1}_L$ would be (almost everywhere) equal to a piecewise constant function, where $L=[g(u_{m-1}),1]\cap K$; since $L$ has positive mesure and its support has empty interior, this is not possible. This thus have a contradiction; accordingly $(f(t))_{0\le t\le 1}$ is indeed free.
In a different direction from Yves Cornulier's suggestion: how about something based on Cauchy kernels as elements of Hardy space? I'm writing this off the top of my head so I might not be taking the most efficient approach.
That is: let ${\mathbb D}$ be the open unit disc in complex plane and let $$ H= \left\{ f : {\mathbb D} \to {\mathbb C} \hbox{ analytic} \colon \sum_{n\geq 0} |a_n|^2 < \infty \right\} $$ where in the above, $f$ has Taylor series $\sum_{n\geq 1} a_nz^n$. Give $H$ the usual norm and inner product, so that (with the obvious abuse of notation) $$ \langle f, g\rangle = \sum_{n\geq 0} a_n \overline{b_n} $$
Take $\displaystyle f_t(z) = \frac{1}{1-tz}$ for $-1/2\leq t \leq 1/2$. Note that $f_t$ has Taylor series $\sum_{n\geq 0} t^n z^n$.
A routine estimate of norms shows that $F: t \mapsto f_t$ is continuous from $[-1/2,1/2]$ to $H$; it is clearly injective. Then $B=F([-1/2,1/2])$ is compact (continuous image of a compact space) and $F:[-1/2,1/2]\to B$ is a homeomorphism (continuous bijection between two compact Hausdorff spaces).
To see that $B$ is linearly independent, note that every $g\in \operatorname{span}(B)$ is a rational function of $z$ with simple poles, all outside $\overline{\mathbb D}$, and the representation of such a function as a linear combination of things in $B$ is unique.
Claim: $F([-1/2,0])$ has dense linear span in $H$.
Proof of claim. Consider the "perp" of this set, i.e. the set of all $h\in H$ such that $\langle h,f_t\rangle = 0$ for all $-1/2\leq t\leq 0$. By the definition of our inner product: if $h(z) = \sum_{n\geq 0} c_n z^n$ then $$ h(t) = \sum_{n\geq 0} c_n t^n=0 \quad\hbox{for all $-1/2\leq t\leq 0$} $$ so that the zero-set of the function $h$ has an accumulation point in the interior of the disc. This forces $h=0$ by standard properties of analytic functions. (Alternatively: just calculate the Taylor coefficients $c_n$ by differentiation to get $c_n=0$ for all $n$.)
Thus $F([-1/2,0])$ has trivial perp, which by basic Hilbert space theory implies it has dense linear span. QED
A similar argument shows that $F([0,1/2])$ also has dense linear span.
So $\overline{\operatorname{span}F([-1/2,0])} \cap \overline{\operatorname{span}F([0,1/2])} = H$, which is quite a lot larger than $\operatorname{span}\{ F(0)\} = {\mathbb C}$.
Remark. The family of functions $f_w(z) = (1- \overline{w}z)^{-1}$, for $w\in {\mathbb D}$ are useful to keep in mind when thinking of how "linearly independent with dense linear span" is a much weaker condition than "forms a (Schauder) basis", when dealing with Banach spaces. They are also the prototypical examples of reproducing kernels for Hilbert spaces of functions.
