Generalization of linear independence to infinite-dimensional spaces Let $n\in\mathbb{N}$, and recall the definition of linear independence on the finite dimensional vector space $\mathbb{R}^n$:

If $\left(v_i\right)_{i=1,..,p}$ a family of $p$ vectors of $\mathbb{R}^n$, it is linearly independent if and only if:
  $$\forall\lambda_1,\lambda_2,..\lambda_p\in\mathbb{R},\hspace{12pt}\left(\sum\limits_{i=1}^{p}\lambda_iv_i=0_{\mathbb{R}^n}\right)\hspace{-3pt}\Longrightarrow\hspace{-3pt}\left(\lambda_1=\lambda_2=..=\lambda_p=0\right)$$

Now, consider the following property linear independence has in the $\mathbb{R}^n$ spaces:

Let the vectors $\left(e_i\right)_{i=1,..,n}$ of $\mathbb{R}^n$ be defined by $e_1=\begin{pmatrix}1\\0\\:\\0\end{pmatrix}, e_2=\begin{pmatrix}0\\1\\:\\0\end{pmatrix}$,.. ,$e_n=\begin{pmatrix}0\\0\\:\\1\end{pmatrix}$
Let $\left(v_i\right)_{i=1,..,n}$ be a linearly independent family of $n$ vectors of $\mathbb{R}^n$ ($n$ being also the dimension of $\mathbb{R}^n$).
A linear map $f$ from $\mathbb{R}^n$ to $\mathbb{R}^n$ that verifies
  $$\forall i=1,..,n, f(e_i)=v_i$$
  is necessarily injective.

This might seem like a fairly obvious property, as well as an understatement: $\left(v_i\right)_{i=1,..,n}$ and $\left(e_i\right)_{i=1,..,n}$ are actually 2 basis of $\mathbb{R}^n$, and $f$ is not only injective but bijective.
Well, the thing is, when generalizing to infinite-dimensional vector spaces, the notions of basis and bijectivity are very tricky to handle, but injectivity and linear independence are much more easy. So I think this weaker statement can be generalized much more easily than the full one.
Now, I wish to find a generalization of the above definition of linear independence valid for countably infinite families of vectors $\left(v_i\right)_{i\in\mathbb{N}}$, that preserves this property, in the following sense:
We now work on the vector space of real sequences $\mathbb{R}^{\mathbb{N}}$, which is the natural infinite-dimensional equivalent of $\mathbb{R}^n$.
The vectors $\left(e_i\right)_{i\in\mathbb{N}}$ of $\mathbb{R}^{\mathbb{N}}$ are now defined by $\forall i\in\mathbb{N}, e_i=\left(\delta_{k,i}\right)_{k\in\mathbb{N}}$, with $\delta$ the kronecker symbol. In other words, $e_i=\begin{pmatrix}0\\:\\1\\:\end{pmatrix}$ is the infinite sequence of $0$'s everywhere exept at the position $i$, which is a $1$.
In this setting, the aforementioned property I wish to be conserved reads (with the proper definition of linear independence):

Let $\left(v_i\right)_{i\in\mathbb{N}}$ be a linearly independent family of vectors of $\mathbb{R}^{\mathbb{N}}$ (i.e sequences).
A linear map $f$ from $\mathbb{R}^{\mathbb{N}}$ to $\mathbb{R}^{\mathbb{N}}$ that verifies
  $$\forall i\in\mathbb{N}, f(e_i)=v_i$$
  is necessarily injective.

I have 2 candidates for defining linear independence so that the above statement is true:

1) $\left(v_i\right)_{i\in\mathbb{N}}$ is linearly independent if and only if every finite subfamily of $\left(v_i\right)_{i\in\mathbb{N}}$ is linearly independent

Or

2) $\left(v_i\right)_{i\in\mathbb{N}}$ is linearly independent if and only if:
  $$\forall\lambda_1,\lambda_2,..\in\mathbb{R},\hspace{12pt}\left(\sum\limits_{i=1}^{\infty}\lambda_iv_i=0_E\right)\hspace{-3pt}\Longrightarrow\hspace{-3pt}\left(\lambda_1=\lambda_2=..=0\right)$$

We can notice that the second definition is stronger than the first : every family that is 2)-linearly independent is 1)-linearly independent, but the converse is not necessarily true.
Does the first definition of linear independence for infinite families preserve the above property, or does only the stronger second definition ? Or maybe none of them ?
 A: Linear independence in  a general vector space is defined by condition 1). There are serious problems with 2) because the infinite sum may not exist. So you simply abandon any attempt to use 2).
Your assertion about injectivity is false. The $e_i$'s do not form a basis (=maximal linearly independent set) for $\mathbb{R}^{n}$ and we can extend them to a maximal linearly independent set in  $\mathbb{R}^{n}$. There will infinitely many new vectors in this extension and you can define a linear map $f$ by assigning arbitrary values to basis vectors. It is clear now that $f$ need not be injective. 
A: Your first claim is not true as $(e_i)_i$ do not form a basis of $\mathbb{R}^\mathbb{N}$ as noted by the other answer. However, you had the right idea. If we really have a basis, we can conclude injectivity, see the theorem below.
I'll make use of the following definitions and notations:

Definition (Linear Independence): A subset $L \subseteq V$ of a vectorspace $V$ is called linearly independent if for every finite subset $S \subseteq L$ we have
  $$\sum_{v\in S} \alpha_v v = 0 \Rightarrow \forall v \in S. \alpha_v = 0$$
Definition (Basis): I'll use Hamel bases where every vector admits a unique representation as a linear combination.
Notation (Lists): The notation $(\ldots)$ refers to unordered lists, which may contain duplicates.

The definition of linear independence is equivalent to your first definition (1). Now let us state and prove

Theorem: Let $f: V \to W$ be a linear function between two vectorspaces. Let $(v_i)_i$ be a basis of $V$ and $(f(v_i))_i$ be linearly independent in $W$. Then $f$ is injective.

Proof: Let $x, y \in W$ with $f(x) = f(y)$. We need to show $x = y$. Since $(v_i)_i$ is a basis, there are representations
$$x = \alpha_{i_1}v_{i_1} + \ldots \alpha_{i_n}v_{i_n}\\
  y = \beta_{j_1}v_{j_1} + \ldots \beta_{j_m}v_{j_m}.
$$
Applying this to $f(x) = f(y)$, we get
$$\alpha_{i_1}f(v_{i_1}) + \ldots + \alpha_{i_n}f(v_{i_n}) = \beta_{j_1}f(v_{j_1}) + \ldots + \beta_{j_m}f(v_{j_m})$$
We now want to apply our linear independence assumption somehow. For that we need something of the form $\ldots = 0$, where $\ldots$ is a linear combination of linearly independent vectors. However, even if we pull the RHS in the equation above to the LHS, we cannot guarantee that some $f(v_{i_k}) = f(v_{j_l})$ for some $k$ and some $l$. Let us consider what happens in those cases. Let's define $A := \{(k, l) : f(v_{i_k}) = f(v_{i_l})\}$ as the indices where the things match up. Let us rewrite the equation:
$$
\begin{align*}
  &\left(\sum_{(k,l) \in A} \alpha_{i_k} f(v_{i_k}) - \beta_{j_l} f(v_{j_l})\right) + \left(\sum_{k \notin A} \alpha_{i_k} f(v_{i_k})\right) - \left(\sum_{l \notin A} \beta_{j_l} f(v_{j_l}) \right) = 0\\
 \Leftrightarrow & \left(\sum_{(k,l) \in A} (\alpha_{i_k} - \beta_{j_l})f(v_{i_k})\right) + \left(\sum_{k \notin A} \alpha_{i_k} f(v_{i_k})\right) - \left(\sum_{l \notin A} \beta_{j_l} f(v_{j_l})\right) = 0
\end{align*}
$$
By $k \notin A$ and $l \notin A$ I mean that these values do not ever occur in their respective position -- $k$ as a left component, $l$ as a right component -- in $A$. You might want to mentally verify that for all values of $k$ (or $l$) the set $A$ either doesn't contain it at all in the previous sense or exactly contains one tuple with it. More than one tuple cannot be the case due to $(f(v_i))_i$ being linearly independent by our assumption.
The LHS now features a linear combination of linearly independent vectors by assumption. Let's continue and apply the definition of linear independence:


*

*$\forall (k,l) \in A. \alpha_{i_k} = \beta_{j_l}$

*$\forall k \notin A. \alpha_{i_k} = 0$

*$\forall l \notin A. \beta_{j_l} = 0$
Hence, we actually have the representations


*

*$x = \sum_{k \in A} \alpha_{i_k} v_{i_k}$

*$y = \sum_{l \in A} \beta_{j_l} v_{j_l}$
So the coefficients agree by our previous results. However, we do not yet know whether the vectors agree themselves. But they do! Namely, for $(k,l) \in A$ we know $f(v_{i_k}) = f(v_{j_l})$ by the very definition of $A$. By linear independence of $(f(v_i))_i$ by our assumption, we can conclude $v_{i_k} = v_{j_l}$. Hence, $x = y$.
