A matrix $M$ is symmetric if it satisfies $M^T=M$. Where ${}^T$ denotes the transpose operation (swapping rows with respective columns and columns with respective rows). For the statement made to be true you actually need the additional condition that all the entries in the matrix are real numbers. I use $u*$ for the complex conjugate of the vector $u$, that is the vector with components conjugate to the respective components of $u$. We write the additional condition $M*=M$. (This type of matrix (real and symmetric) is part of a set of operators called Hermitian operators which are defined by the condition $M{*}^T=M$).

Suppose that $Mv=\lambda v$ (that $v$ is an eigenvector of a matrix $M$ with eigenvector $\lambda$) then since $v$ is not the zero vector we have $|v{|}^2=v{*}^{T}v>0$ (where $\left|v\right|$ is the length (or norm) of the vector $v$):

$\lambda =\lambda v{*}^{T}v/(v{*}^{T}v)=v{*}^{T}Mv/(v{*}^{T}v)=v{*}^{T}M{*}^{T}v/(v{*}^{T}v)=(Mv){*}^T/(v{*}^{T}v)=(\lambda v){*}^{T}v/(v{*}^{T}v)=\lambda *(v{*}^{T}v)/(v{*}^{T}v)=\lambda *$

So if $M$ has an eigenvalue it is real. It is not necessarily true that $M$ (when it is $n$ x $n$) has $n$ eigenvalues (consider the identity matrix which has only one eigenvalue). It is true that $M$ has at least one eigenvalue and that it has no more than $n$, to prove this you need to assume (because proving it is rather tricky) the fundemental theroem of algebra which states that every polynomial has a root in the set of complex numbers. By induction you can show easily enough that this implies that a polynomial of degree (largest power) $n$ has no more than $n$ distinct roots in $C$.

The function

$\chi (x)=\det (M-xI)$

(where $I$ is the ( $n$ x $n$) identity matrix) has roots which are precisely the eigenvalues of $M$. $\det (M-xI)=0$ only when $M-xI$ is not invertible which happens and only happens when there exists a vector $v$ such that $(M-xI)v=0$ (such that $Mv=xv$) (This might need a little more explaning).

Your argument is identical to William's. Since the map $(x)\to x$ is a field isomorphism, we can consider $(x)$ the same as $x$, where $x$ is a scalar.

If you are still unhappy about this, consider:

$\lambda v{*}^{T}v=v{*}^T(\lambda v)=v{*}^{T}Mv=v{*}^{T}M{*}^{T}v=(Mv){*}^{T}v=\lambda *v{*}^{T}v$

i.e. $(\lambda -\lambda *)v{*}^{T}v=0$

since $v{*}^{T}v$ is nonzero, we have $\lambda =\lambda *$.