Proof that vector space $C([a,b])$ of all functions from $[a,b]\to \mathbb{R}$ is infinite dimensional? Let me first express how I understand an infinite dimensional vector space. 

A vector space V is infinite dimensional if $\forall n\in \mathbb{N} (\{e_1,\dots , e_n\} \text{ linearly independent }\Rightarrow \{e_1 , \dots , e_{n+1}\}\text{ linearly independent })$.

Having the above definition in mind, I try to prove that the vector space $C([a,b])$ of all functions from $[a,b]\to \mathbb{R}$ is infinite dimensional, with basis the set of polynomials.
Attempt at a proof: Let $n\in \mathbb{N}$ be arbitrary, and assume that $\{x , x^2 , \dots , x^n\}$ is linearly independent. My goal is to show that $\{x,x^2,\dots , x^{n+1}\}$ is linearly independent. To do that, assume that $\sum_{k=1}^{n+1}b_kx^k=0$, and now I need to show that for all $k$, $b_k=0$. My feeling is that this should proceed by contradiction, namely I should assume that there exists $k$ such that $b_k\neq 0$... 
I am missing the step from here on, how do I proceed. Also is the definition of infinite dimensionality correct? Is this how one proves that a vector space is infinite dimensional?
 A: A vector space is infinite-dimensional if there does not exist a finite set that spans it -- is generally taken as the definition. More precisely, one defines a vector space to be finite-dimensional if there is a finite set that spans it and then an infinite-dimensional space is then merely defined to be a space that is not finite-dimensional.
EDIT- In response to your comment, the span of a set of vectors is the set of all linear combinations of those vectors. So, for instance the vector space of complex numbers over the field of real numbers is finite-dimensional because it is spanned by the finite set $\{1,i\}$
A: First of all you forgot to include $1$ in your set $\{x , x^2 , \dots , x^n\}$.
Also the assumption that  $\{1, x , x^2 , \dots , x^n\}$. is linearly independent is not necessary to show that  to show that  $\{1, x , x^2 , \dots , x^{n+1}\}$ is linearly independent. 
Let $C_1+C_2x+...+C_{n+2}x^{n+1}=0$.
Plog $x=0$ and you get $C_1=0$. Now differentiate and evaluate at $x=0$ and you will get $C_2=0$.
Continue and you will get all the coefficients equal $0$
Therefore you have an infinite set {${1, x , x^2 , ... , x^n, ...}$} of linearly independent continuous functions.     
A: Your definition at the moment doesn't quite make sense, since you haven't specified what the $e_n$'s are. It's also incorrect - what happens if $e_1 = 0$? Since there's quite a few different equivalent definitions, I will try and state several here.
To clarify notation, fix a vector space $V.$ We say a finite subset $F = \{e_1,\dots,e_n\}\subset V$ is linearly independent if for any choice of scalars $\lambda_1,\dots,\lambda_n,$ we have,
$$ \sum_{i=1}^n \lambda_ie_i = 0 \quad \iff \quad \lambda_1=\dots=\lambda_n = 0.$$
A general subset $F \subset V$ is linearly independent if every finite subset $F_0 \subset F$ is linearly independent. This is equivalent to the assertion that for all $n \in \mathbb N,$ $x_1,\dots, x_n \in F$ and $\lambda_1,\dots,\lambda_n$ scalars, $\sum_{i=1}^n \lambda_ie_i = 0$ if and only if each $\lambda_i=0.$
Also a finite subset $F = \{e_1,\dots,e_n\} \subset V$ spans $V$ if for all $x \in V,$ there are scalars $\lambda_1,\dots,\lambda_n$ such that $x = \sum_{i=1}^n \lambda_ie_i.$ Similarly a general subset $F \subset V$ spans $V$ if for all $x \in V,$ there is $n \in \mathbb N,$ $x_1,\dots,x_n \in F$ and scalars $\lambda_1,\dots,\lambda_n$ such that $ x = \sum_{i=1}^n\lambda_ix_i.$
Theorem The following are equivalent.


*

*There exists an infinite subset $F \subset V$ such that $F$ is linearly independent.

*There exists a countably infinite subset $F = \{e_n\}_{n \in \mathbb N} \subset V$ such that $F$ is linearly independent.

*There exists a countably infinite subset $F = \{e_n\}_{n \in \mathbb N} \subset V$ such that for all $n \in \mathbb N,$ the subset $F_n = \{e_1,\dots,e_n\}$ is linearly independent.

*There exists a countably infinite subset $F = \{e_n\}_{n \in \mathbb N} \subset V$ such that $e_1 \neq 0$ and for all $n \in \mathbb N$ (starting at $n=1$),
$$ \{e_1,\dots,e_n\} \text { is linearly independent } \implies \{e_1,\dots,e_{n+1}\} \text{ is linearly independent.} $$

*There does not exist a finite subset $F \subset V$ spanning $V.$

*(*) There exists a infinite subset $F \subset V$ which is both linearly independent and spanning.
If any of the above equivalent conditions holds, we say $V$ is infinite dimensional.
(*) Strictly speaking, the equivalence with (6) is not only non-trivial, but is equivalent the axiom of choice. It is commonly used however since it allows one to define the dimension of $V$ to be the cardinality of the basis $F,$ but I digress... Also on the note of choice, I will be using countable choice freely.
Proof: We will first show the series of implications,
$$ (\mathrm{iv}) \implies (\mathrm{iii}) \implies (\mathrm{ii}) \implies (\mathrm{i}) \implies (\mathrm{iv}). $$
For $(\mathrm{iv}) \implies (\mathrm{iii})$ this is simply the principle of mathematical induction. We note $\{e_1\}$ is linearly independent provided $e_1 \neq 0,$ which establishes the base case.
For $(\mathrm{iii})\implies(\mathrm{ii}),$ we note that if $F \subset{e_n}_{n \in \mathbb N}$ is finite, then $n = \max F$ exists. So $F \subset \{e_1,\dots,e_n\}$ from which the implication follows. The implication $(\mathrm{ii})\implies (\mathrm{i})$ is trivial since $\{e_n\}_{n \in \mathbb N}$ is an infinite linearly independent subset. The final implication $(\mathrm{i}) \implies (\mathrm{iv})$ holds by noting said infinite linearly independent subset $F$ contains a countable subset $\{e_n\}_{n \in \mathbb N} \subset F.$ This is automatically linearly independent, so we can verify $(\mathrm{iv}).$
Finally $(\mathrm{i}) \iff (\mathrm{v})$ is equivalent to showing $\neg(\mathrm{i}) \iff \neg(\mathrm{v}),$ where no infinite linearly independent set exists if and only if a finite spanning set exists. This will require some results from finite dimensional linear algebra, which I will assume. The latter implication will also need a bit of choice also.
If a finite spanning set $F$ exists, then we know any set $G \subset V$ with $|G| > |F|$ (set cardinality) must be linearly dependent. This gives one direction.
For the converse, assume that any infinite subset $F \subset V$ is linearly independent. The strategy will be to start with a linearly independent set inductively add linearly independent vectors to our set until it spans the space. Indeed we will fix $x_1 \in V\setminus\{0\}$ and set $F_1 = \{x_1\}.$ Given $F_n = \{x_1,\dots,x_n\}$ is linearly independent, if $\mathrm{span}\ F_n \neq V$ then there exists $x_{n+1} \in V$ such that $F_{n+1} = \{x_1,\dots,x_{n+1}\}$ is linearly independent. So we continue inductively (note this uses dependent choice). If this process doesn't terminate, we will get an infinite subset $\{x_1,x_2,\dots\}$ which is linearly independent, contradicting our original assumption. So the process eventually terminates to give a finite spanning set.
For reasons mentioned above, I will skip the equivalence with $(\mathrm{vi}).$

For your specific problem I'll refer you to Mohammad Riazi-Kermani's answer, which gives an inductive proof (so it's an example of using $(\mathrm{iv})$).
A: Your supposed definition of infinite dimensional vector space doesn't make sense. A vector space $V$ is infinite dimensional if no finite subset of $V$ generates $V$. This is equivalent to the assertion that $V$ has an infinite subset which is linearly independent.
The space $\mathcal{C}\bigl([a,b]\bigr)$ is infinite dimensional because the set $\{1,x,x^2,x^3,\ldots\}$ is linearly independent.
