Suppose $V$ is a finite dimensional vector space equipped with a symmetric bilinear form $\langle \cdot ,\cdot \rangle$. Then prove the subset $V^+$ is a subset, where $V^+$ is the set of all $v$ such that if $v$ is nonzero, then $\langle v,v\rangle > 0$.
I showed that $0$ is vacuously in the set, and that $$v\in V^+ \implies cv\in V^+, \quad c\in K.$$ I don’t know where to start to show that $$v,w\in V^+ \implies v+w\in V^+.$$
An inner product (scalar product) is always positive definite, so we have $\langle v,v\rangle > 0$ for any $v\neq 0$ and $\langle 0,0\rangle = 0$. Hence what you describe as $V^+$ is in fact all of $V$ and of course $V^+=V$ is a subspace of itself.
Please comment if I misunderstood your question, it seems a bit trivial.
For an arbitrary symmetric bilinear form, the statement is false: Consider the symmetric bilinear form $b(v,w)=v_1w_1-v_2w_2$ on $\mathbb R^2$. This yields $$ V^+ = \{(0,0)\}\cup\{\, (v_1,v_2) \in\mathbb R^2 \,:\, |v_1|>|v_2|\,\}, $$ which is not a subspace. For example, $(2,1)$ and $(-2,1)$ are in $V^+$ but their sum $(0,2)$ isn't.