Given finitely many points in a vector space $V$, is there a basis such that the first coordinate of each point is distinct? Suppose I have some $n$-dimensional vector space $V$ and a finite collection of $m$ distinct points $v_1,\dotsc, v_m\in V$. Is there a basis of $V$ such that the first coordinate of each $v_i$ is distinct?
This obviously fails when the base field is finite, but my intuition over $\mathbb{R}^n$ has convinced me it's true when the base field is infinite. I'm having a hard time proving it though. Any suggestions?
 A: You can think of this as being related to the following fun result. A short proof is given by Steve D here.

Fact: If $V$ is a vector space over an infinite field $F$ (the vector space can be finite or infinite dimensional), then $V$ cannot be expressed as the union of finitely many proper subspaces.

Returning to your question, it is convenient to first consider the finite-dimensional case.

Claim: Let $V$ be a finite-dimensional vector space over an infinite field $F$. Let $D$ be a finite set of nonzero vectors in $V$. Then, there exists a hyperplane $W \subset V$ such that $D \cap W = \varnothing$.
Proof: Let $n$ be the dimension of $V$. We construct recursively subspaces $\{0\} = W_0 \subset W_1 \subset \ldots \subset W_{n-1}$ such that each is disjoint with $D$ and $\dim W_k =k$. Suppose that $W_k$ is already constructed for $k < n-1$. By the fact quoted above, $V$ is not the union of the subspaces $W_k + F \cdot d$ as $d$ ranges over the elements of $D$. Let $v \in V$ be such that $v \notin W_k + F \cdot d$ for all $d \in D$ and put $W_{k+1} = W_k + F\cdot v$.
Corollary: Let $V$ be a finite-dimensional vector space over an infinite field $F$. Let $S$ be a finite subset of $V$. Then, there exists a linear functional $\varphi : V \to F$ taking distinct values on the elements of $S$.
Proof: Let $D$ be the set of differences $x-y$ where $x$ and $y$ are distinct elements of $S$. So, $D$ is a finite set of nonzero vectors in $V$. By the above claim, there is a hyperplane $W$ such that $D \cap W = \varnothing$. Since no two elements of $S$ are equal modulo $W$, they all correspond to distinct elements in the 1-dimensional quotient $V/W$. Choosing an identification of $V/W$ with $F$ gives the desired functional.

This takes care of the finite-dimensional version of your question as follows: choose $\varphi$ as above and extend it to a basis $\varphi,\varphi_2,\ldots,\varphi_n$ for the dual space $V^*$. Since we are in the finite-dimensional case, these functionals correspond to a basis $e_1,\ldots,e_n$ for $V$ such that the coordinate function of $e_i$ is exactly $\varphi_i$.
On the other hand, it's easy to see to see the finite-dimensional version of your problem implies the infinite-dimensional version. Given a finite subset $S$ of an infinite-dimensional vector space $V$ over an infinite field $F$, we can find a basis for the finite subspace $\mathrm{span}(S)$ such that the elements of $S$ all get different first coordinates, and then extend arbitrarily to a basis for all of $V$.
A: Let $V$ be a vector space of dimension $n$ over a finite field.
Lemma: Given a finite set of vectors $F$ there exists $W\leq V$ of dimension $n-1$ with $W\cap F=\varnothing$.
We prove every maximal subspace with $W\cap F=\varnothing$ must have dimension $n-1$.
Suppose not: Let $W$ be a maximal subset with $W\cap F$ and let $U$ be a subspace of dimension $2$ with $W\cap U=\varnothing$.
Notice that for each $u\in U$ we have $f-\alpha u \in W$ for some non-zero scalar $\alpha$ and $f\in F$. Given $f\in F$ let $X_f$ be the set of vectors $u\in U$ such that $f-\alpha u\in W$ for some scalar $\alpha$. We conclude $\bigcup\limits_{f\in F}X_f=U$.
If we define $Y_f$ as the subspace spanned by $X_f$ then it should also be clear that $\bigcup\limits_{f\in F} {Y_f}=U$. This implies that $Y_f=U$ for some $f$ (because a finite union of proper subspaces of a vector space over an infinite field cannot be the entire vector space).
Therefore we can obtain $u_1$ and $u_2$ linearly independent vectors in $X_f$. Notice that we have $f-\alpha u_1\in W$ and $f-\beta u_2\in W$, which implies $\beta u_2-\alpha u_1\in W$, but we also have $\beta u_2-\alpha u_1\in U$. Which implies $\beta u_2-\alpha u_1=0$. A contradiction. Hence the lemma is proved.

Applying the lemma to our particular case is trivial. Let $v_1,v_2\dots v_m$ be the vectors and consider the set $u_1,u_2,\dots u_k$ constisting of all values $v_i-v_j$.
By the lemma there exists a subspace of dimension $n-1$ $W$ such that $u_i\not\in W$ for all $1\leq i \leq k$.
Let $b_2,b_3\dots b_n$ be a basis for $W$.
Then any extension $b_1,b_2\dots b_n$ does the trick.
