We first prove that for any $$ x_0\in <!--\; \in \; -->[a,b]$$ and $$ \mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ there exists some integer $$ N\in <!--\; \in \; -->{\mathbb{N}}_1$$ and $$ r\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ such that $\[ |x-<!--\; -\; -->x_0|\le <!--\; \le \; -->r⟹<!--\; ⟹\; -->\left|g_N\left(x\right)\right|\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->} \]$ where $$ g_n\left(x\right)=f\left(x\right)-<!--\; -\; -->f_n\left(x\right)$$

This follows from a few facts, we first start with continuity of $$ f,f_n$$, specifically at $$ x_0$$ and therefore using $$ \mathrm{ϵ<!--\; ϵ\; -->}$$ we obtain some $$ {\mathrm{\delta <!--\; \delta \; -->}}_1\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ such that $$ f\left(B(x_0,{\mathrm{\delta <!--\; \delta \; -->}}_1)\right)\subseteq <!--\; \subseteq \; -->B(g_n\left(x_0\right),\mathrm{ϵ<!--\; ϵ\; -->})$$, similarly we have a $$ {\mathrm{\delta <!--\; \delta \; -->}}_2\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ such that $$ f\left(B(x_0,{\mathrm{\delta <!--\; \delta \; -->}}_2)\right)\subseteq <!--\; \subseteq \; -->B(f\left(x_0\right),\mathrm{ϵ<!--\; ϵ\; -->})$$, then also we know that since $$ f_n\stackrel{\mathrm{pw}}{\to <!--\; \to \; -->}f$$ then since it's true at $$ x_0$$ and using $$ \mathrm{ϵ<!--\; ϵ\; -->}$$ we obtain some $$ K\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that for all $$ n\in <!--\; \in \; -->{\mathbb{N}}_K$$ we have $$ \left|f_n\left(x_0\right)\right|-<!--\; -\; -->f\left(x_0\right)\le <!--\; \le \; -->\mathrm{ϵ<!--\; ϵ\; -->}$$

Now we combine using the triangle inequality, so let $$ \mathrm{ϵ<!--\; ϵ\; -->}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ , then let $$ r=min({\mathrm{\delta <!--\; \delta \; -->}}_1,{\mathrm{\delta <!--\; \delta \; -->}}_2)$$ so that we obtain the inequalities mentioned in the previous paragraph, and we get some $$ K\in <!--\; \in \; -->{\mathbb{N}}_1$$ as mentioned in the previous paragrah, then

With that out of the way suppose for the sake of contradiction that $$ f_n$$ doesn't convege uniformly to $$ f$$, since $$ f-<!--\; -\; -->f_n$$ is eventually bounded then this is equivalent to $\[ \underset{n\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}\Vert <!--\; \Vert \; -->f-<!--\; -\; -->f_n{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}=d\in <!--\; \in \; -->{\mathbb{R}}^{+} \]$

Since $$ g_n$$ is continuous as the sum of continuous functions and has compact domain then by EVT, we know that it attains its max and min, and therefore there exists a $$ x_k\in <!--\; \in \; -->[a,b]$$ such that $$ \Vert <!--\; \Vert \; -->f-<!--\; -\; -->f_k{\Vert <!--\; \Vert \; -->}_{\mathrm{\infty <!--\; \infty \; -->}}=g_k\left(x_k\right)$$, so we have $$ \underset{k\to <!--\; \to \; -->\mathrm{\infty <!--\; \infty \; -->}}{lim}{g}_k\left(x_k\right)=d$$, and note that this means that for any $$ {\mathrm{ϵ<!--\; ϵ\; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\in <!--\; \in \; -->{\mathbb{R}}^{+}$$ there is some $$ J\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that for all $$ j\in <!--\; \in \; -->{\mathbb{N}}_J$$ we have $$ |g_j\left(x_j\right)-<!--\; -\; -->d|<<!--\; <\; -->{\mathrm{ϵ<!--\; ϵ\; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}$$ which implies that $$ d-<!--\; -\; -->{\mathrm{ϵ<!--\; ϵ\; -->}}^{\mathrm{\prime <!--\; \prime \; -->}}\le <!--\; \le \; -->g_j\left(x_j\right)$$.

Since $$ x_k$$ is in a compact set it has a convergent subsequence $$ x_{n_k}\to <!--\; \to \; -->x\in <!--\; \in \; -->[a,b]$$. Now recall the fact we proved initially, by using $$ d-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$ we know that there is some $$ N\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that $$ g_N\left(x\right)<<!--\; <\; -->d-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}$$, now at the same time the inequality discussed in the previous paragraph also holds on the subsequence (this is because if a sequence goes somewhere, then any subsequence must also go there), namely we have some $$ J\in <!--\; \in \; -->{\mathbb{N}}_1$$ such that for all $$ j\in <!--\; \in \; -->{\mathbb{N}}_J$$ $\[ d-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}\le <!--\; \le \; -->g_{n_j}\left(x_{n_j}\right) \]$ but then there is some $$ i$$ such that $$ n_i><!--\; >\; -->max(N,J)$$ and therefore $$ g_i<<!--\; <\; -->g_N$$ and so chaining inequalities we have $\[ d-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->}\le <!--\; \le \; -->g_{n_i}\left(x_{x_i}\right)\le <!--\; \le \; -->g_N\left(x\right)\le <!--\; \le \; -->d-<!--\; -\; -->\mathrm{ϵ<!--\; ϵ\; -->} \]$ which is a contradiction, and therefore $$ f_n\stackrel{\mathrm{unif}}{\to <!--\; \to \; -->}f$$