Suppose we have an infinite dimensional vector space $V$, also $\beta$ is a subset of $V$. So, $\beta$ is a basis for $V$ if and only if for every non-zero vector $v\in V$, $v=c_1u_1+c_2u_2+......+c_nu_n$ where $c_i$'s are non-zero scalars and $u_i$'s are some vectors in $\beta$.
First, we say $\beta$ is a basis of $V$ which must be infinite as $V$ is infinite dimensional.
$(*)$ Now, $v\in V$, so $v$ must be in a span of some $S$ where $S$ is some finite subset of $\beta$.
There could be other such S's whose span contain $v$, so I consider the intersection of all of them, and, say their intersection contains vectors $u_1,u_2,u_3,....u_n$.
The reason I say 'intersection' is to make the scalars non-zero, as in, consider $P(R)$ which of course is infinite dimensional, $x^2+1\in span(\{1,x,x^2\})$ and $x^2+1\in span(\{1,x^2,x^3\})$, when I write $x^2+1$ with respect to any of the sets mentioned, some scalars have to be zero, but if I consider their intersection, then they are not.
So we've found some vectors such that $v=c_1u_1+c_2u_2+......+c_nu_n$ where $c_i$'s are non-zero.For uniqueness of these vectors we could say that if there is any other representation of $v$, say $v=a_1v_1+a_2v_2+......+a_nv_n$, then $c_1u_1+c_2u_2+......+c_nu_n+(-a_1)v_1+-(a_2)v_2+......+(-a_n)v_n=0$
Now all these $u_i$'s and $v_j$'s are in $\beta$ and hence must be linearly independent, thus all $a_i$'s,$c_i$'s are indeed $0$, thus $v=0$, but that's not possible since we are not considering the zero vector and hence the representation is unique.
For the converse part, i.e. assuming each vector has such a unique representation, could be done like the way we did in the case of finite dimensional vectors, can we ?
Like, since each vector in $V$ can be written uniquely in the form of some vectors of $\beta$, those 'some' vectors are indeed linearly independent and since this is true for each and every vector of $V$, thus $\beta$ must be a basis for $V$.
Is that correct ?
I am specially skeptical for the $(*)$ statement !