Real valued continuous functions on [a,b] form a vector space with respect to usual addition and multiplication by scalars. 
Real valued continuous functions on $[a,b]$ form a vector space with respect to usual addition and multiplication by scalars.

Please help to show a proof.  I think it would be a laborious one.  Is it really?
 A: It is quite laborious, but it's not too difficult. In my experiences, students tend to struggle with this kind of question particularly simply because it's so straightforward, they think they're not doing it properly, and it doesn't feel like they've actually proven anything.
Here, let me get you started. First, I'll address closure of the operations. Then I'll prove associativity of vector addition, then I'll establish the additive identity, and I'll leave you to do the rest.
Suppose $f, g \in C[a, b]$ and $\lambda \in \mathbb{R}$. The definitions of addition and scalar multiplication are respectively,
\begin{align}
(f + g)(x) &:= f(x) + g(x) &\forall x \in [a, b] \\
(\lambda f)(x) &:= \lambda f(x) &\forall x \in [a, b]
\end{align}
By the algebra of continuous functions, $f + g$ and $\lambda f$ are both continuous on $[a, b]$.
Suppose $f, g, h$ are continuous on $[a, b]$. We must establish that $(f + g) + h = f + (g + h)$. That is, we must establish, for all $x \in [a, b]$,
$$((f + g) + h)(x) = (f + (g + h))(x).$$
Suppose $x \in [a, b]$. Using the definition of vector addition, as well as the associativity of real numbers, we get
\begin{align*}
((f + g) + h)(x) &= (f + g)(x) + h(x) &\ldots \text{ definition} \\
&= (f(x) + g(x)) + h(x) &\ldots \text{ definition} \\
&= f(x) + (g(x) + h(x)) &\ldots \text{ associativity} \\
&= f(x) + (g + h)(x) &\ldots \text{ definition} \\
&= (f + (g + h))(x) &\ldots \text{ definition}
\end{align*}
This holds for any $x \in [a, b]$, so $(f + g) + h = f + (g + h)$.
Finally, to establish the identity $\mathbf{0}$, we must specify what it is. We let $\mathbf{0}$ be the function that maps any $x \in [a, b]$ to $0$. This is a constant function, hence it is continuous. We now must prove it is an identity, that is, $\mathbf{0} + f = f$ for any $f$.
(In a wider sense, in order to establish $\mathbf{0}$ is truly an identity, we would also need to prove $f + \mathbf{0} = f$, but you'll be proving commutativity later, which will make this redundant.)
Suppose $f$ is continuous on $[a, b]$ and $x \in [a, b]$. Then,
$$(\mathbf{0} + f)(x) = \mathbf{0}(x) + f(x) = 0 + f(x) = f(x).$$
Therefore $\mathbf{0} + f = f$.
Now you can establish the rest! You don't necessarily need as much detail as I'm using while writing it out, but you do need to cover all the axioms.
A: If you have to prove it without knowing anything about function spaces, @Theo Bendit has pointed out what you have to do. However, if you already know that $$\mathcal{V}=\{f:\mathbb R\rightarrow\mathbb R\}$$ is a vector space, it is enough to show that $$\mathcal{W}=\{f:\mathbb R\rightarrow\mathbb R~|~f~\text{continuous}\}$$ is a subspace of of $\mathcal{V}$.

To prove that:


*

*Show that the $f\equiv 0\in\mathcal{W}$

*Let $f,g\in\mathcal{W}$ and $a\in\mathbb R$. Show that this implies $f+a\cdot g\in\mathcal{W}$.

