I am trying to prove the following theorem:
A Bilinear form $B$ on a finite-dimensional vector space $V$ is symmetric if it is diagonalizable.
Here is my attempt at a proof.
Suppose $B$ is diagonalizable. Since $B$ is a bilinear form on a finite-dimensional vector space for any basis $S=\{s_1,\dots,s_n\}$ of $V$ we can find a matrix $A = (c_{ij})$ such that for all $v,w\in V$:
$$ B(v,w) = [v]_S^T A [w]_S $$
Since $B$ is diagonalizable choose $S' = \{s_1',\dots,s_n'\}$ to be a basis for $V$ that diagonalizes $A$. We can write $[v]_{S'} = \left(\begin{array}{c} a_1 \\ \vdots \\ a_n\end{array}\right)$, and $[w]_{S'} = \left(\begin{array}{c} b_1 \\ \vdots \\ b_n\end{array}\right)$, and $A=(\alpha_{ij})$ where $\alpha_{ij} = \left\{ \begin{array}{cc} c_{ij} & i=j \\ 0 & \text{otherwise} \end{array}\right.$.
Then if we compute $B(v,w)$ we obtain:
$$ \begin{align*} B(v,w) &= [v]_{S'}^TA[w]_{S'} \\ &= \left( \begin{array}{ccc} a_1 & \cdots & a_2\end{array}\right)A\left( \begin{array}{c} b_1 \\ \vdots \\ b_n\end{array}\right) \\ &= \left( \begin{array}{ccc} c_{11}a_1 & \cdots & c_{nn}a_n\end{array}\right)\left( \begin{array}{c} b_1 \\ \vdots \\ b_n\end{array}\right) \\ &= \sum_{i=1}^n c_{ii}a_ib_i \\ &= \sum_{i=1}^n c_{ii}b_ia_i \\ &= \left( \begin{array}{ccc} c_{11}b_1 & \cdots & c_{nn}b_n\end{array}\right)\left( \begin{array}{c} a_1 \\ \vdots \\ a_n\end{array}\right) \\ &= \left(\begin{array}{ccc} b_1 & \cdots b_n \end{array}\right)A\left( \begin{array}{c} a_1 \\ \vdots \\ a_n\end{array}\right) = B(w,v) \end{align*} $$
$\blacksquare$
In class we proved it by induction on the dimension of $V$, however, this seems like a more direct and intuitive approach. Is it valid?
Yes, but that's perhaps not the beginner way to do it. What you're doing is that you consider the bilinear form without a basis - the beginner variant consider it in some basis and having it represented as a matrix and then introduce transformation matrices. Your proof becomes simpler, but somewhat more "advanced".
As for using induction that's often more of a formality in this proof. When you use ellipsis it's actually a more "visual" way to describe it and when used in proofs it's a simplified way to describe something that would have been proved using induction otherwise.