The problem statement is as follows. We are given a function $f:[0,1]\times [0,1] \to \mathbb{R}$. For all fixed $x$, $y\mapsto f(x,y)$ is continuous, and likewise, for all fixed $y$, $x\mapsto f(x,y)$ is continuous. We are also given that the set defined by $$\mathcal{E} = \{x\mapsto f(x,y): y\in[0,1]\}$$ is equicontinuous. How would I show that $f(x,y)$ is continuous?
Progress so far: Assume that $f(x,y)$ is discontinuous at $(x_0,y_0)$. Fix $x_0$ and consider a sequence $(y_n) \to y$; define $f_n = f(x_0, y_n)$ for each element $y_n \in (y_n)$. I want to show that $f_n\rightrightarrows f$. However, since $y_n \to y$, by continuity of $y\mapsto f(x,y)$ we have $$|y_n - y| < \delta \implies |f(y_n) - f(y)| < \epsilon$$ This implies that there exists $N\in \mathbb{N}$ such that $\forall n \geq N$ and $y\in [0,1]$, $|f(y_n) - f(y)| < \epsilon$, implying $f_n \rightrightarrows f$.
Since the uniform limit of continuous functions is continuous, this would imply that $f(x_0,y)$ is continuous for all $y \in [0,1]$. Contradiction, since $f$ was assumed to be discontinuous at $(x_0, y_0)$.
I think this works, but I'm skeptical about the part of my proof that establishes the uniform convergence of the $f_n$ to $f$. Any clarification would be much appreciated.
Note: I've also tried restricting the convergent sequence of $y_n$ to rationals and applying the Arzela-Ascoli propagation theorem, but I'm not sure that works since we're not guaranteed that the convergence involves all elements in $Q\cap [0,1]$.