Let $U'$ be a subset of $U$ by choosing each triangle with the probability $p$ independently at random. Then the expected number of triangles in $U'$ is $mp$.

A sequence of 6 distinct points $(x_1, x_2, \dots, x_6)$ is called a bad configuration if all 6 triangles $x_1x_2x_3, x_2x_3x_4, \dots, x_4x_5x_6, x_5x_6x_1, x_6x_1x_2$ belong to $U$.

The number of bad configurations in $U$ is at most $m(m-1)(3!)^2 \le 36m^2$, because it is less than or equal to the number of selecting two triangles, selecting $x_1, x_2, x_3$ from the first triangle, and selecting $x_4, x_5, x_6$ from the second triangle.

If $(x_1, x_2, \dots, x_6)$ is a bad configuration, then $(x_i, x_{i+1}, \dots, x_{i+6})$ and $(x_i, x_{i-1}, \dots, x_{i-6})$ are bad configurations and so $6 \cdot 2 = 12$ bad configurations form a bunch. Thus the number of bunches of bad configurations in $U$ is at most $36m^2/12 = 3m^2$.

The probability that a fixed bad configuration of $U$ is contained in $U'$ is $p^6$ and therefore the expected number of bunches of bad configurations in $U'$ is at most $3m^2p^6$.

Thus, the expected number of triangles in $U'$ minus the number of bunches of bad configurations in $U'$ is at least $mp - 3m^2p^6$.

Take $p = c m^{-1/5}$. Then
$$mp - 3m^2p^6 \geq c m^{6/5} - 3c^6 m^2 m^{-6/5} = (c - 3c^6)m^{4/5}.$$
By taking $c = 1/2$, we deduce $mp - 3m^2p^6 \geq (\frac{1}{2} - \frac{3}{64})m^{4/5} \geq 0.45m^{4/5}$.

Therefore there exists a subset $U'$ of $U$ such that the number of triangles in $U'$ minus the number of bunches of bad configurations in $U'$ is at least $0.45m^{4/5}$. Let $W$ be a subset of $U'$ by discarding one triangle in each bunch of bad configurations. Then $W$ contains at least $0.45m^{4/5}$ triangles and no bad configurations, as desired.