You are right in most of what you say, and you seem to be missing a very tiny, though important, basic theorem from linear algebra:
Definition: Let $\,V:=V_{\Bbb F}\,$ be a vector space . A subspace $\,H\,$ is called a hyperplane if it is maximal subspace of $\,V\,$, i.e.: $\,H\,$ is a hyperplace iff for every
$x\in V\setminus H\;\;,\;\;V=Span\{H,x\}=:\langle\,H\,,\,x\,\rangle$
If $\,\dim_{\Bbb F}V=n<\infty\,$ , the above condition is equivalent with $\,\dim_{\Bbb F}H=n-1\,$
Theorem:
(1)${}\,\,\,\,\,{}$For any $\,f\in V^*:=Hom_{\Bbb F}(V,\Bbb F)\,$ , either $\,f\,$ is trivial or else $\,f\,$ is onto.
(2)${}{}\,\,\,\,{}$ A subspace $\,H\leq V\,$ is a hyperplane iff $\,H=\ker f\,$ , for some $\,0\neq f\in V^*\,$.
Thus, as you can see, the intersection of two hyperplanes indeed is a non-trivial subspace of the space (with $\,\dim V= n=2\,$ being the only non-trivial exception), and the condition of infinte dimension is not important in this case.
Proof of (2) (Highlights):
Suppose $\,H\leq V\,$ is a hyperplane and let $\,\{x_i\}_{i\in I}\,$ be a basis of $\,H\,$. Take now $\,v\notin Span\{x_i\}_{i\in I}\,$ , then $\,B:=\{x_i\,,\,v\}_{i\in I}\,$ is a basis of $\, V\,$ (why?), and define $\,f:V\to\Bbb F\,$ by extending linearly the function
$f:B\to\Bbb F\;\;\;,\;\;f(y):=\begin{cases}0&\text{if}\,\,\, y=x_i\,\,,\,\,\text{for some}\,\,i \in I\\{}\\1&\text{if}\,\,y=v\end{cases}$
Check now that $\,H=\ker f$
If $\,H=\ker f\,$ , for some $\,0\neq f\in V^*\,$ , and if $\,f(x)=t\neq 0\,\,,\,\,x\in V\,$ , then $\,V=\langle\,H,x\,\rangle\,$ , because: let $\,v\in V\,$, then: if $\,f(v)=0\,$ then $\,v\in\ker f\,$, otherwise
$f(v)=k\neq 0\Longrightarrow \,\exists! z\in\Bbb F\,\,s.t.\,\,k=zt\Longrightarrow f(v)=k=zt=zf(x)=f(zx)\in\langle x\rangle$
Try to finish and round up the argument now.