Let $\{X_i\}$ a family of topologcial vector space(equipped with topology $\mathcal{T_i}$)
let $X = \bigcup_{i = 1}^\infty X_i$ equipped with inductive topology w.r.t $\{(X_i,\mathcal{T}_i,I_i)\}$(the finest topology which makes $I_i:X_i \to X$ continuous where $I_i$ is the identity map into)
Prove if for each $i$ we have $V\bigcap X_i$ is neighborhood near 0 in topological vector space $X_i$
Then $V$ is a neighborhood near 0 in topological space $X$
I think what you're asking comes down to the following result. It has nothing to do with topological vector spaces.
Proposition: Let $\mathcal T$ be the inductive topology on $X$ coming from the spaces $(X_i, \mathcal T_i)$, and let $\mathcal T'$ be the topology on $X$ given by the condition that $V \subseteq X$ is open if and only if $V \cap X_i$ is open in $X_i$ for each $i$. Then $\mathcal T = \mathcal T'$.
Proof: By definition, $\mathcal T$ is the largest topology on $X$ for which each inclusion $X_i \subset X$ is continuous. In other words, if $\mathcal T''$ is a topology on $X$ for which each inclusion $X_i \rightarrow X$ is continuous, then $\mathcal T'' \subseteq \mathcal T$. Since each inclusion $X_i \rightarrow X$ is clearly continuous for $\mathcal T'$, we have $\mathcal T' \subseteq \mathcal T$.
On the other hand, if $V \subseteq X$ belongs to $\mathcal T$, then each $V \cap X_i$ is open in $X_i$, and hence $V \in \mathcal T'$. Thus $\mathcal T \subseteq \mathcal T'$.