Answer:
Here I will assume that [tex]V[/tex] is the vector space of functions [tex]f:\mathbb{R}\rightarrow\mathbb{R}[/tex]. If [tex]V[/tex] is the space of continuous functions, there will be no changes in the proof, because polynomial functions are also continuous.
Let us write [tex]P[/tex] for the set of polynomials with real coefficients. As we want to prove that [tex]P[/tex] is a subspace of [tex]V[/tex] we need to check three conditions:
First: [tex]P[/tex] is a non empty set, in particular [tex]0\in P[/tex]. This is not difficult, because the function identically 0 is a polynomial. Then, [tex]P[/tex] is not empty and contains the 0 vector of [tex]V[/tex].
Second: The addition of two elements of [tex]P[/tex] stays in [tex]P[/tex]. Consider two arbitrary polynomials [tex]p[/tex] and [tex]q[/tex]. Then,
[tex]p(x) = a_0+a_1x+\cdots+a_nx^n[/tex]
[tex]q(x) = b_0+b_1x+\cdots+b_mx^m[/tex].
This means that [tex]p[/tex] is a polynomial of degree [tex]n[/tex] with coefficients [tex]a_k[/tex], and [tex]q[/tex] is a polynomial of degree [tex]m[/tex] with coefficients [tex]b_k[/tex]. With out lose of generality assume that [tex]m\geq n[/tex].
Now, notice that
[tex]p(x)+q(x) = (a_0+b_0) + (a_1+b_1)x + \cdots+(a_n+b_n)x^n +b_{n+1}x^{n+1} +\cdots +b_mx^m[/tex]
which is a polynomial too.
Hence, [tex]p+q\in P[/tex].
Third: The multiplication by a scalar stays in [tex]P[/tex].
Consider an arbitrary polynomial
[tex]p(x) = a_0+a_1x+\cdots+a_nx^n[/tex]
and a real number [tex]\alpha[/tex].
Then,
[tex]\alpha p(x) = (\alpha a_0)+(\alpha a_1)x+\cdots+(\alpha a_n)x^n[/tex]
which is a polynomial.
Hence, [tex]\alpha p\in P[/tex].
Therefore, [tex]P[/tex] is a subspace of the vector space [tex]V[/tex].
