Answer: $c = \frac{2}{5}$.
We say that a simple graph with $n$ vertices is c-good if it is triangle-free and the degree of any vertex is strictly bigger than $cn$. A length 5 cycle is triangle-free and the degree of any vertex is 2. For $c < \frac{2}{5}$, this graph is c-good, but not bipartite.

Now we show that for $c = \frac{2}{5}$, all c-good graphs are bipartite. Let $G$ be a triangle-free graph which is not bipartite. We prove that it has a vertex with degree $\le \frac{2n}{5}$.

By Euler's theorem, $G$ contains an odd cycle. Let
$$C = \{v_1, v_2, \dots, v_m\}$$
be an odd cycle of minimum length. Since $G$ is triangle free, we have $m \ge 5$.

First, note that vertices in $C$ are connected iff they have adjacent indices (modulo $m$). Otherwise, an edge connecting two non-adjacent vertices $v_i$ and $v_j$ divides $C$ into two cycles of combined length $m + 2$:
$$\{v_i, v_{i-1}, \dots, v_{j+1}, v_j\}, \quad \{v_i, v_{i+1}, \dots, v_{j-1}, v_j\}.$$
One of these cycles will have an odd length strictly less than $m$, which contradicts the minimality of $m$.

Similarly, a vertex not in $C$ can connect to at most two vertices in $C$. Indeed, suppose that $u \notin C$ is connected to $v_r$, $v_s$, $v_t \in C$ with $1 \le r < s < t \le m$, and consider the three cycles of combined length $m + 6$:
$$\{u, v_r, v_{r+1}, \dots, v_s\}, \quad \{u, v_s, v_{s+1}, \dots, v_t\}, \quad \{u, v_t, v_{t+1}, \dots, v_r\}.$$
One of these cycles will have an odd length and the other two will have length at least four, since $G$ is triangle free. Therefore, the odd cycle will have length at most $m + 6 - 4 - 4 = m - 2$. This again contradicts the minimality of $m$.

Finally, let $d(u)$ denote the degree of a vertex $u \in G$ and let $d(u, C)$ denote the number of edges from $u$ to a vertex in $C$. From above, we have $d(u, C) \le 2$ for any $u \in G$. Thus we have
$$\sum_{v \in C} d(v) = \sum_{u \in G} d(u, C) \le 2n.$$

It follows that there is a vertex $v$ in $C$ such that $d(v) \le \frac{2n}{m} \le \frac{2n}{5}$. This completes the solution.