Let $V$ be a finite dimensional real vector space and $f$ a symmetric bilinear form on $V\times V$. Define the signature of $f$ $\;s(f)$ to be $a-b$ where $a$ is the number of 1s on the diagonal and $b$ is the number of -1s on the diagonal when $f$ is represented by a diagonal matrix (and the basis vectors are normalised such that diagonal entries are 0,1 or -1). Note $a$ is the largest dimension of a subspace on which $f$ is positive definite and similarly for $b$.
Let $c,d\in V$ be such that $f(c,c)=0$ and $f(c,d)=1$. Show $\exists W\le V$ such that $V$ is the direct sum of span$(c,d)$ and $W$ and that $s(f|_W)=s(f)$.
What I have shown so far:
If $K\le V$ is such that $f(k,v)=0$ for all $k\in K$ and $v\in V$ then $f'(v+K,v+K):=f(v,v)$ defines a symmetric bilinear form on $V/K$ and $s(f')=s(f)$
$c,d$ are linearly independent and the possible signatures of $f|_{span(c,d)}$ are 0,1,-1.
$U=\{v\in V: f(c,v)=f(d,v)=0\}$ has that the sum $U+$span$(c,d)$ is direct.
$V$ equals the direct sum $U+$span$(c,d)$
Update:
Motivated by Marc van Leeuwen's comment it is indeed easily shown that $s(f|_{span(c,d)})=0$.
Combining our results we also have (apologies I do not know how to format a block matrix, please do edit if you are able) there exists a basis $(v_1,v_2)$ for span$(c,d)$ such that for any basis $(v_3,...,v_n)$ of $U$ the matrix representation of $f$ in the basis $(v_1,...,v_n)$ for $V$ is a block matrix with top left block $ \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix} $ , zeros in the remainder of the first two rows and columns, and bottom right $(n-2)\times (n-2)$ block $ \begin{pmatrix} f|_U \end{pmatrix} $
From the updates the result is obvious, since we can take a basis $(v_3,...,v_n)$ of $U$ such that $(f|_U)$ is diagonal with 1,-1,0 on the diagonal, and hence $s(f)=s(f|_{span(c,d)})+s(f|_U)=s(f|_U)$