Reducing independent set to triangle-free subgraph
Let $\langle G=(V, E), k \rangle$ be an instance of Independent Set, and call $n=|V|$.
Let $N$ be a set of $n+1$ new vertices (not in $V$), and construct a new graph $G' = (V', E')$ where $V' = V \cup N$ and $E' = E \cup (N \times V)$.

If $G$ has an Independent Set $S$ of size at least at least $k$ then $G'$ has a Triangle-Free set $S'$ of size at least $n+k+1$.

Select $S' = S \cup N$.
To see that the subgraph of $G'$ induced by $S'$ is triangle free consider any triple $(a,b,c)$ of distinct vertices in $S'$.
Clearly, if more than one of $a$, $b$, and $c$ is in $N$, the triple cannot induce a triangle since $G'$ has no edges between vertices in $N$. On the other hand, if at least two distinct vertices $u,v \in \{a, b, c\}$ are in $S$, then $(u,v) \not \in E$ and hence $(u,v) \not\in E'$.

If $G$ has a Triangle-Free set $S'$ of size at least $n+k+1$ then $G'$ has an Independent Set $S$ of size at least at least $k$.

Let $S=S' \setminus N$.
Since $|S| \ge n+k+1$ and $|N|=n+1$, the cardinality of $S$ is at least $k$.
To see that $S$ is an Independent Set of $G$ notice that $S' \cap N \neq \emptyset$ and let $v \in S' \cap N$. Let $a$ and $b$ be any two distinct vertices in $S$. Since the subgraph of $G'$ induced by $S'$ is triangle free and $\{(v,a), (v,b) \} \subseteq E'$, we must have $(a,b) \not\in E' \supseteq E$.