Diagonalizability of symmetric bilinear forms over fields of characteristic $2$ 
Theorem: 
A symmetric bilinear form $H$ on a finite dimension vector space $V$
  over a field $\mathbb{F}$, where it is not of characteristic two, is
  diagonalizable.

Proof:

By induction on $\text{dim} \; V=n$.
(induction base) $n=0$. Then it is trivial.
(induction hypothesis) Assume the above statement holds for all
  bilinear forms on vector spaces of dimension $n-1$.
(inductive step) Suppose the space $\text{dim} \; V=n$. If the
  bilinear form $H=0$, then it is trivial. Hence, assume that $H\neq0$,
  then there exists $z \in V$ such that $H(z,z) \neq 0$. Let $W =
\operatorname{span}\{z\}^\perp$ and we have $V = W \oplus
\operatorname{span}\{z\}$. As $\text{dim} \; W=n-1$, the theorem holds
  for this space and there is a basis $\beta = \{v_1, ..., v_{n-1}\}$
  where $H$ is diagonal. Then, by extending the basis we have $\gamma = \beta \cup \{z\} \subset V$. Then we have: $H(v_{i},z) = 0$ for all
  $i=0, ..., n-1$. Hence, this implies that there exists a basis $\gamma
\subset V$ such that the matrix corresponding to $H$ is diagonal.


Why is the field of characteristic two excluded? 
 A: Over $\mathbb{F}_2$, you can find symmetric bilinear forms which are not diagonalizable. Take, for instance, the bilinear form $B\colon{\mathbb{F}_2}^2\times{\mathbb{F}_2}^2\longrightarrow\mathbb{F}_2$ defined by$$B\bigl((x_1,x_2),(y_1,y_2)\bigr)=x_1y_1+x_1y_2+x_2y_1+x_2y_2.$$In ${\mathbb{F}_2}^2$ there are only three bases (without caring about the order of the vectors): $\bigl\{(1,0),(0,1)\bigr\}$, $\bigl\{(1,0),(1,1)\bigr\}$, and $\bigl\{(0,1),(1,1)\bigr\}$. You can check that the matrix of $B$ with respect to each one of these bases is not diagonal.
Note: This is a corrected version of my answer, made after Omnomnomnom telling, in the comments, that I was using the wrong concept of diagonalizable.
A: 
... Hence, assume that $H\neq0$, then there exists $z \in V$ such that $H(z,z) \neq 0$. ...

This is not true in characteristic $2$. Let $\Bbb F$ be any field of that characteristic, and, for example, take $V = \Bbb F^2$ and the bilinear form $H$ with matrix representation
$$[H] = \pmatrix{0&1\\1&0}$$
with respect to the standard basis; then, the quadratic form $Q_H : {\bf x} \mapsto H({\bf x}, {\bf x})$ is the zero form. Indeed, we can see directly that $H$ is not diagonalizable: Computing directly for any $P \in \textrm{GL}(2, \Bbb F)$ gives
$$P^T [H] P = (\det P) [H],$$
which is not diagonal.
Put another way, it is not true in characteristic $2$ that the map $H \mapsto Q_H$ is injective. In other characteristics it is injective, as we can recover $H$ from $Q$ via the Polarization Identity
$$H({\bf x}, {\bf y}) = \tfrac{1}{4}[Q_H({\bf x} + {\bf y}) - Q_H({\bf x} - {\bf y})] ,$$
but in characteristic $2$ one cannot divide by $4$ (which in that setting coincides with $0$).
Remark The above facts imply (since the spaces of symmetric bilinear forms on $V$ and quadratic forms on $V$ both have dimension $\frac{1}{2} (\dim V)(\dim V + 1)$) that (only) in characteristic $2$ there are quadratic forms that are not induced by symmetric bilinear forms (that is, are not in the image of the map $H \mapsto Q_H$); the simplest example is $V = \Bbb F^2$ and
$$\pmatrix{x\\y} \mapsto x y .$$
