Function is not continuous as a multivariable function but continuous as a single variable one

Question

It's known that a multivariable function can be continuous in one variable, but doesn't necessary need to be also continuous as a multivariable one (see $\frac{xy}{x^2+y^2}$ as a counter example). Anyway, I can't see what's wrong with the following proof, in which it seems that the claim is actually true:
Assuming $f(x,y) \stackrel{y \rightarrow 0}{\rightarrow}f(x,0)$ for every $x$, and $f(x,y) \stackrel{x \rightarrow 0}{\rightarrow}f(0,y)$ for every $y$: for every $\epsilon > 0$, there are $\delta_1 > 0, \delta_2 > 0 $ such that for every $0<|x|<\delta_1, 0<|y|<\delta_2$: $|f(x,y)-f(0,0)| < |f(x,y)-f(0,y)| + |f(0,y)-f(0,0)| < \frac{\epsilon}{2}+ \frac{\epsilon}{2}<\epsilon$.
I used the given continuity of $f$ according to $x$ and $y$ separetly, and the triangle inequality. I fail to see what is flawed in my line of thinking, and would like it if someone could point out what is the problem.

Answer 1

The assertion$$\lim_{y\to0}f(x,y)=f(x,0)\tag1$$lacks quantifiers. What is $x$? If what you mean is that, for each $x$, $(1)$ holds, then what this means is that, for each $x$ and for each $\varepsilon>0$, there is some $\delta>0$ (depending on $x$ and on $\varepsilon$) such that$$|y|<\delta\implies\bigl|f(x,y)-f(x,0)\bigr|<\varepsilon.$$So, the $\delta_1$ from your proof does not exist. To be more precise, there is one for each individual $x$.