Here's a picture that might help. A visual way of understanding $\delta$-$\epsilon$ arguments is by starting with a $\delta$-sized area in the domain, projecting up to the function, and then back onto an $\epsilon$-sized area in the range.
With the function $f(x) = x$, there is a bounded ratio between the size of the area I feed in and the size of the area I get out. Not so with $f(x) = x^2$! Look how I feed in small areas and get out large areas for large values of $x$. This is why we say that $f(x) = x$ is uniformly continuous, but $f(x) = x^2$ is not uniformly continuous on $\mathbb{R}$. There's no way to globally (i.e. independent of $x$) control the size of the image of $f(x) = x^2$ by controlling the size of the domain.
In general,if $X,Y$ are any spaces, a function $f:X\times X\to Y$ may have the property that the functions $f^{(1)}_a:X\to Y$ and $f^{(2)}_a:X\to Y$ are continuous for every $a\in X,$ where $f^{(1)}_a(x)=f(a,x)$ and $f^{(2)}_a(x)=f(x,a),$ and yet $f$ may still not be continuous.
What you need to show is that when $u,v\in \mathbb R$ with $u<v$ then $d^{-1}(u,v)=\{(a,b)\in X:u <d(a,b)<v\}$ is open in the space $X\times X.$
An open set in $X\times X$ is a union of a family of sets of the form $C\times D$ where $C,D$ are open balls in $X.$ So it suffices to show that if $(a,b)\in d^{-1}(u,v)$ then $C\times D\subset d^{-1}(u,v)$ for some open balls $C,D$ of $X$ with $(a,b)\in C\times D.$ That is, with $a\in C$ and $b\in D.$
For $d(a,b)\in (u,v),$ let $w=\min (d(a,b)-u,\; v-d(a,b)).$ We have $w>0$ so $a\in B_d(a,w/2)$ and $b\in B_d(b,w/2)$.
For $a'\in B_d(a,w/2)$ and $b'\in B_d(b,w/2)$ we have $$\text {(i)} \quad d(a',b')\leq d(a',a)+d(a,b)+d(b,b')<w/2+d(a,b)+w/2\leq v$$ and we have $$\text {(ii) }\quad d(a,a')+d(a',b')+d(b',b)\geq d(a,b)$$ which implies $$\text {(ii') }\quad d(a',b')\geq d(a,b)-d(a,a')-d(b',b)>d(a,b)-w/2-w/2\geq u.$$ Therefore $$(a,b)\in d^{-1}(u,v)\implies (a,b)\in B_d(a,w/2)\times B_d(b,w/2)\subset d^{-1}(u,v).$$
Remark: For an example of an $f$ of my first paragraph, let $X=Y=\mathbb R,$ let $f(0,0)=0$ and let $f(x,y)=xy/(x^2+y^2)$ when $x^2+y^2\ne 0.$ If $x\ne 0$ then $f(x,x)=1/2$ so $\lim_{x\to 0^+}f(x,x)=1/2\ne f(0,0)$. So $f$ is not continuous at $(0,0).$
Best Answer
Yes. Example: Take the metric where $d(x,x) = 0$ and $d(x,y)=1$ for $x\neq y$ and the Euclid metric. In the first one all functions are continuous and in the second you can find non-continuous functions