I proved that every associative algebra with multiplication $\ast$ turn into a Lie algebra by a Lie bracket derived from $\ast$.
$\textbf{ My question:}$
Can an arbitrary vector space $V$ over a field $F$ be turned into an algebra over $F$ (associative or nonassociative algebra) by defining a nontrivial bilinear map $\ast\colon V\times V\rightarrow V$ as a product?
If yes, how? If no, why?
To define an algebra on $V$ with basis $\{e_i\}_{i\in I}$, you need structure constants $$ e_ie_j=\sum_k\Gamma_{ij}^ke_k $$ that satsify a few axioms (e.g. for associativity or commutativity), then extend everything linearly.
One can come up with examples off the top of one's head, e.g. $e_ie_j=\delta_{ij}e_i$ which is the "coordinate-wise" algebra on $V\cong\bigoplus_{i\in I}Ke_i$ if the base field is $K$. However, this isn't an algebra over $K$ if $V$ is infinite dimensional (the infinite sum $1=\sum_ie_i$ that you would want doesn't exist without defining some topology). I believe you can fix this by "reserving" some basis element $e_0$ to act as the identity $e_0e_i=e_ie_0=e_i$.
If you don't care about it being an "algebra" in the sense that there's a map $K\hookrightarrow V$, then the zero product or the product defined by $e_ie_j=\delta_{i,j}e_i$ is bilinear (and associative and commutative).
In functional analysis, when an identity doesn't exist, one can be adjoined or the algebra can be enlarged. For instance, the space of compactly supported continuous functions $C_c(\mathbb{R})$ can be embedded in the space of bounded continuous functions.