No. Much like how the axiom of choice is not involved in the statement "a subset of a finite set is finite".
But to your specific question, every infinite set contains finite sets of arbitrarily large cardinality. Otherwise, $A$ is an infinite set, and $B$ is a maximal finite subset, so $B\subsetneq A$, since one is finite and the other is not; but then take any $a\in A\setminus B$, and consider $B\cup\{a\}$, which is a strictly larger finite subset of $A$.
So in particular, if $M$ is linearly independent and infinite, it contains arbitrarily large subsets which are, well, linearly independent.
The only one thing you need to make sure, though, is that when you consider your infinite set of vectors, it is not "a list" which somehow implies that you think about it as indexed by $\Bbb N$. It is perfectly possible to have a field (and a finite dimensional vector space over it) which is infinite, but has no countably infinite subset.
Nevertheless, the question is about infinite subsets. Not infinite lists. So this is a minor point on your choice of language.
Yes you are correct. But note that we are talking about a Hamel basis here. Every Hilbert space also has a more useful notion of basis, called orthonormal basis, which is a set of orthonormal elements in the Hilbert space with dense span. Such a basis can be countable, for example $l^2(\Bbb{N})$ has orthonormal basis the Dirac functions on the natural numbers.
Best Answer
This is for sure not true. Take any separable Hilbert space with a countable Hilbert basis $\mathcal B = \{b_i \mid i \in I\}$ and $ j \in I$. Then $\mathcal B \setminus \{b_j\}$ can't be a basis while it is a countable set of linearly independent vectors.