Uncountability of basis of $\mathbb R^{\mathbb N}$ Given vector space $V$ over $\mathbb R$ such that the elements of $V$ are infinite-tuples. How to show that any basis of it is uncountable?

Take any almost disjoint family $\mathcal A$ of infinite subsets of $\mathbb N$ with cardinality $2^{\aleph_0}$.
Construction of such set is given here.
I.e. for any two set $A,B\in\mathcal A$ the intersection $A\cap B$ is finite.
Notice that 
$$\{\chi_A; A\in\mathcal A\}$$
is a subset of $\mathbb R^{\mathbb N}$ which has cardinality $2^{\aleph_0}$.
We will show that this set is linearly independent. This implies that the base must have cardinality at least $2^{\aleph_0}$. (Since every independent set is contained in a basis - this can be shown using Zorn lemma. You can find the proof in many places, for example these notes on applications of Zorn lemma by Keith Conrad.)
Suppose that, on the contrary,
$$\chi_A=\sum_{i\in F} c_i\chi_{A_i}$$
for some finite set $F$ and $A,A_i\in\mathcal A$ (where  $A_i\ne A$ for $i\in F$). The set $P=A\setminus \bigcup\limits_{i\in F}(A\cap A_i)$ is infinite.
For any $n\in P$ we have $\chi_A(n)=1$ and $\sum\limits_{i\in F} c_i\chi_{A_i}(n)=0$. So the above equality cannot hold. 

You can find a proof about dimension of the space $\mathbb R^{\mathbb R}$ (together with some basic facts about Hamel bases) here: Does $\mathbb{R}^\mathbb{R}$ have a basis?
In fact, it can be shown that already smaller spaces must have dimension $2^{\aleph_0}$, see Cardinality of a Hamel basis of $\ell_1(\mathbb{R})$.
