Let $V$ be some $k$-vector space, $v,w\in V$. How can we see that
$$\gamma(v) = \gamma(w) \quad\forall \gamma\in V^* \quad \implies \quad v = w?$$
This question can be rephrased as
Prove that the pairing $V^*\times V\to k,$ $(\gamma,v)\mapsto\gamma(v)$ is non-degenerate (at least on the second argument).
It is painfully obvious if you allow choice: Just take any old basis. BAM.
But can we prove this without using a basis? I don't know, and my advisor couldn't come up with a proof either.
No.
If $k$ is any field, then it is consistent that there is a vector space $V$ over $k$ such that: (1) $V$ is nonzero; (2) the only endomorphisms of $V$ are scalar multiplication.
These conditions in conjunction imply that there are no non-zero functionals from $V$ to $k$. In particular, you cannot separate points in $V$ using functionals from $V^*$.
This was my masters thesis, building on an originally lesser known proof by Läuchli from the 1960s (I believe it was his Ph.D. thesis).
Additionally, one can obtain similar results from theorems about Banach spaces and the Hahn–Banach theorem's failure. Specifically, if every set of reals has the Baire property, then $\ell^\infty/c_0$ has a trivial dual.