Convergence of sequence of functions along "diagonal". Consider a sequence of functions $\{f_n\}$ such that $f_n \to f$ uniformly on a compact subset $K \subset C(\mathbb{C})$. Now consider a family of sequences of functions $\{g_{n,m}\}$ such that $g_{n,m} \to f_m$ uniformly on $K$ for all $n \in \mathbb{N}$. I would like to conclude that there exists some sequence of elements in the set $\{g_{n,m}\}$ which converges uniformly to $f$ on $K$. I have tried considering the diagonal $\{g_{n,n}\}$. This led me naturally to an $\frac{\epsilon}{3}$ argument, namely: 
$$\|f-g_{n,n}\| \leq \|f-f_m\| + \|f_m-g_{n,m}\| + \|g_{n,m}-g_{n,n}\|$$
Fix $\epsilon>0$, then take $m$ large enough for the first term and then $n$ large enough for the second. I can then take $n$ larger if need be to make the third term small enough, but I cannot vary $m$ without affecting the initial choice of $n$, so there doesn't seem to be any way to control the size of the third term. 
I believe I would need something like the following: 
If $N>0$ is given, then we have $\|f_m-g_{n,m}\|<\epsilon_m$ for $n>N$. If we knew that $\{\epsilon_m\}$ was bounded, then the result would follow. I'm not sure if this is true though. It seems plausible, since $f_n \to f$, but all of my proof attempts thus far have failed. Perhaps there is a counterexample? 
If there is a counterexample, are there any additional conditions would make it true besides the aforementioned?
 A: By definition of uniform convergence, 
\begin{align}
f_n &\rightrightarrows f & \iff &
& \forall \, \varepsilon_{f} &> 0 & \exists\; N_{{\varepsilon}_{f}}
%\underbrace{N\varepsilon}_{\qquad=N_{\varepsilon\,}\left(\varepsilon,\,f\right)}
&\in\mathbb N: 
&\forall \,n & \ge N_{{\varepsilon}_{f}}
&\left\lVert f_n\left(x\right)-f\left(x\right)\right\rVert &< \varepsilon_{f}
\end{align}
Similarly, assuming $n\in \mathbb N$ is fixed, we have
\begin{align}
g_{m,n} &\rightrightarrows f_n & \iff &
& \forall \, \epsilon_{f_n} &> 0& \exists\, M_{\epsilon_{f_n}} &\in \mathbb N : 
& \forall \,m & \ge M_{\epsilon_{f_n}} 
& \left\lVert g_{m,n} \left(x\right) - f_n\left(x\right)\right\rVert & < \epsilon_{f_n}
\end{align}
Combining the last two equations we can write
\begin{align}
\left\lVert g_{m,n} - f \right\rVert = 
\left\lVert g_{m,n} - f_n + f_n - f\right\rVert 
\leq \left\lVert g_{m,n} - f_n\right\rVert + \left\lVert f_n  - f \right\rVert
\end{align}
Let us choose an arbitrary positive constant $\boldsymbol \varepsilon >0$.
Without loss of generality  we can choose $\varepsilon_f$ and $\epsilon_{f_n}$ such that $\epsilon_{f_n} + \varepsilon_f\leq \boldsymbol \varepsilon$.
Choosing $m,n\in\mathbb N$ such that $m \ge M_{\epsilon_{f_n}}$ and $n \ge N_{{\varepsilon}_{f}}$ respectively, we get
\begin{align}
\left\lVert g_{m,n} - f \right\rVert
\leq \left\lVert g_{m,n} - f_n\right\rVert + \left\lVert f_n  - f \right\rVert
\leq \epsilon_{f_n} + \varepsilon_f \leq \boldsymbol \varepsilon
\end{align}
whenever $m,n>\boldsymbol{N}:=\max\left(M_{\epsilon_{f_n}},N_{{\varepsilon}_{f}}\right)$.
Thus we have shown uniform convergence  $g_{m,n} \rightrightarrows f$.

Let $\big\lbrace g_{\boldsymbol k}\big\rbrace\subseteq \big\lbrace g_{m,n}\big\rbrace$ denote subsequence of $\big\lbrace g_{m,n}\big\rbrace$, where $g_{\boldsymbol k} := g_{\boldsymbol k,\boldsymbol k} =g_{m,n}\big\rvert_{\substack{m=\boldsymbol k\\n=\boldsymbol k}}$  such that $g_{\boldsymbol k} \rightrightarrows f$, you can choose $\boldsymbol k\ge\boldsymbol K :=\max\left(M_{\epsilon_{f_n}}, N_{\varepsilon_f}\right)$.
Then
\begin{align}
\forall \, \boldsymbol\varepsilon &> 0
& \exists\, \epsilon_{f_n},{\varepsilon}_{f} &> 0 :&\forall \,\boldsymbol k  &\ge \max\left(M_{\epsilon_{f_n}}, N_{\varepsilon_f}\right)
& \left\lVert g_{\boldsymbol k} - f_n\right\rVert 
& \leq \epsilon_{f_n} + \varepsilon_f = \boldsymbol \varepsilon
\end{align}
In other words, sequence $\left\lbrace g_{\boldsymbol k}\right\rbrace$ converges to function $f$ uniformly:
\begin{align}
\bbox[1ex, border:solid 2pt #e10000]{\begin{aligned}
\forall \, \boldsymbol\varepsilon > 0 \quad\exists\, \boldsymbol K \in \mathbb N^+\!:\quad\forall \, \boldsymbol k > \boldsymbol K \quad \left\lVert g_{\boldsymbol k} - f_n\right\rVert & \leq\boldsymbol \varepsilon
& \implies && \left\lbrace g_{\boldsymbol k}\right\rbrace &\rightrightarrows f
\end{aligned}}
\end{align}
