I've rewritten the following proof from Friedberg, Insel, and Spence's Linear Algebra. I am only unclear on the boldfaced implication. Why is the conclusion following "therefore" true? I understand the latter half of the same sentence is due to the rank-nullity theorem.
Theorem (Every Symmetric B.F. is Diagonalizable): Let $V$ be a finite-dimensional vector space over a field $F$, where $\text{char} (F)\neq 2.$ Then, every symmetric bilinear form on $V$ is diagonalizable.
We prove by induction on $n=\dim V.$ If $n=1,$ then every element of $\mathcal B(V)$ (this is the set of all bilinear forms on $V$) is diagonalizable. Assume for induction that the theorem holds for any number less than $n\in\mathbb{N}.$ It remains to show that the theorem holds in the $n$-case. If $H$ is $0_{\mathcal B(V)},$ then it is diagonalizable; so, assume $H\neq 0_{\mathcal B(V)}.$ By Lemma 3, there exists a nonzero $x\in V$ such that $H(x,x)\neq 0.$ Recall from the previous proposition that $L_x$ is a linear function and $L_x(x)=H(x,x)\neq 0.$ Therefore, rank$(L_x)=1$ which implies that nullity$(L_x)=n-1$ by the rank-nullity theorem. Thereby, $H|_{N(L_x)}$ is a symmetric bilinear form on a vector space of dimension $n-1.$ Applying the induction hypothesis, we have that there exists an ordered basis $(v_1,\dots,v_{n-1})$ for $N(L_x)$ such that $H(v_i,v_j)=0,$ $i\neq j.$ Notice that $x\notin N(L_x)$ and since $N(L_x)$ is a subspace of $V$, we can extend the basis of $N(L_x)$ to one for $V$ by adding $x$. That is, $\beta = (v_1,\dots,v_{n-1},x)$ is linearly independent and $|\beta|=n=\dim V$ implies that $\beta$ is an ordered basis of $V$. Since $v_i\in N(L_x)$ and $H$ is symmetric, $H(v_i,x)=H(x,v_i)=0$ for any $1\leq i \leq n-1.$ Therefore, $\Psi_\beta (H)$ is diagonal which implies $H$ is diagonalizable.
$L_x : V \to F$. Since $F$ is one-dimensional, the rank of $L_x$ is either $1$ or $0$. It is not zero since it is not the zero map, so its rank is $1$.