I'll only deal with real-valued functions, but the complex case is easy to get from this.
A positive operator on $C(X)$ is continuous because its norm satisfies $\|T\| = \|T(1_X)\|_{\infty}$:
Writing $1_X$ for the constant function $x \mapsto 1$, we have for all $g \in C(X)$ that $ -\|g\|_\infty \cdot 1_X \leq g \leq \|g\|_\infty \cdot 1_X, $ so positivity of $T$ yields $ -\|g\|_\infty T(1_X) \leq T(g) \leq \|g\|_\infty T(1_X). $
and therefore $\|T(g)\|_\infty \leq \|g\|_\infty \cdot \|T(1_X)\|_\infty$. This shows that $\|T\| \leq \|T(1_X)\|_\infty$ and taking $g = 1_X$ we see that a positive operator must have $\|T\| = \|T(1_X)\|_\infty$.
On the other hand, we are given that $\|T_n(f) - f\|_\infty \to 0$, where $f$ is assumed to be strictly positive. By compactness of $X$ and continuity of $f$ there is $x_0 \in X$ such that $0 \lt f(x_0) \leq f(x)$ for all $x \in X$. This tells us that $f(x_0) 1_X \leq f$ and by positivity of the operators $T_n$ we conclude that $0 \leq f(x_0) T_n(1_X) \leq T_n(f)$ whence $\|T_n\| = \|T_n(1_X)\|_\infty \leq \frac{1}{f(x_0)} \|T_n(f)\|_\infty$.
However, since $\|T_nf - f\|_\infty \to 0$, there exists a constant $C$ such that $\|T_n(f)\|_\infty \leq C$ for all $n$. In conclusion, $\|T_n\| \leq \frac{C}{f(x_0)}$ for all $n$, which is equicontinuity of the family of linear operators $T_n$.