Let's have a function $f(x,y)$ defined on a continuous $M\subset \mathbb{R}$. For this function $f(x,\cdot )$ $f( \cdot ,y)$ are continuous as functions of $y$ with $x$ fixed, $x$ with $y$ fixed respectively for all $(x,y)\in M$. Also one of the functions, say $f(x,\cdot)$ is monotonous for all $x \in M$. Then this function is continuous on $M$ as a function of two variables.
For $\epsilon >0$ and a point $(x_0,y_0)\in M$ we need to find a neighbourhood of this point $U$ so that $|f(x,y)-f(x_0,y_0)|<\epsilon$ for $(x,y)\in U$. My attempt works with
\begin{align*}
|f(x,y)-f(x_0,y_0)|&=|f(x,y)-f(x_0,y)+f(x_0,y)-f(x_0,y_0)|\\
&\le |f(x,y)-f(x_0,y)|+|f(x_0,y)-f(x_0,y_0)|
\end{align*}
and we could be able to make the right hand side small enough based on the continuity of $f$ as a function of one variable. But I am not sure about this. It also does not make use of the monotonity of $f(x,\cdot)$ (is it really needed?).
Could you give examples of functions that are continuous as one-variable functions, i.e. $f(x,\cdot )$ $f( \cdot ,y)$ but are not continuous as functions of two variables?
Take the function from $\mathbb{R}^2$ to $\mathbb{R}$, that to $x,y$ associates $f(x,y)=\frac{xy}{x^2+y^2}$ if $x,y$ is distinct from $(0,0)$ and $0$ otherwise. The question of continuity is at (0,0), because elsewhere the denominator is obviously non zero. Now if you fix either $x$ or $y$ to be $0$, you will get a continuous function always equal to $0$.
To prove this function is discontinuous as a function of two variables, you just write $x=rcos(\theta )$ and $y=rsin(\theta )$. Then your function will be $f(x,y)=cos(\theta )sin(\theta )$. Then you just make $r$ go to zero and $\theta$ vary so that your function has no limit at $(0,0)$ even though $(rcos(\theta ),rsin(\theta ))$ will approach 0 as $r$ will go to $0$.