Adding a slighly different proof (same idea though), mainly for the sake of documenting it for myself and so I won't forget the theorem:
First let $t \in K$ be fixed. Note that monotonicity of the sequence $f_j(t)$ and its convergence to $f(t)$ implies that $f_j(t) \leq f(t)$.
For $\delta >0$, we define $U_\delta(t) = \left\{\tilde{t} \in K: |\tilde{t}-t|< \delta\right\}$.
Now pick $\varepsilon > 0$, and pick $N_t$ and $\delta_t > 0$ such that:
- $\left|f_n(t) - f(t)\right| \leq \frac{\varepsilon}{3} \; \forall \; n\geq N_t$
- $\left|f_{N_t}(\tilde{t}) - f_{N_t}(t)\right| \leq \frac{\varepsilon}{3} \; \forall \; \tilde{t} \in U_{\delta_t}(t)$
- $\left|f(\tilde{t}) - f(t)\right| \leq \frac{\varepsilon}{3} \; \forall \; \tilde{t} \in U_{\delta_t}(t)$
These exist by the pointwise convergence and the continuity of $f_{N_t}$ and $f$.
It now holds for every $n \geq N_t$:
$ \sup_{\tilde{t} \in U_{\delta_t}(t)} | f(\tilde{t}) - f_n(\tilde{t})|=\sup_{\tilde{t} \in U_{\delta_t}(t)} f(\tilde{t})-f_n(\tilde{t}) \leq \sup_{\tilde{t} \in U_{\delta_t}(t)} f(\tilde{t}) - f_{N_t}(\tilde{t}) \leq f(t) - f_{N_t}(t) + 2\frac{\varepsilon}{3} \leq \varepsilon $
Finally, notice that $\displaystyle\bigcup_{t \in K} U_{\delta_t}(t) \supset K$ is an open cover of $K$. Compactness yields $t_1, \dotsc, t_r \in K$ such that $\displaystyle\bigcup_{i \in \{1,\dotsc,r\}} U_{\delta_{t_i}}(t_i) \supset K$
Now just take $N:= \max\{N_{t_1},\dotsc, N_{t_r}\}$ to finish the proof.