Let $f:D\subset_{open} \mathbb{R^2} \rightarrow \mathbb{R}$ continuous, $\frac{ \partial f}{\partial x}=f_x$ exists and is continuous on $D$ and let $K\subset D$ a closed rectangle. Then, by the Mean Value Theorem $f$ is Lipschitz continuous on $K$ w.r.t. the second variable $x$, i.e.: $|f(t,x_1)-f(t,x_2)| \leq L_K|x_1-x_2| \quad \forall (t,x_1),(t,x_2)\in K$, where $L_K=max_{(t,x)\in K}|f_x(t,x)|$.
I'm not sure how we came to this conclusions since the Mean Value Theorem for real valued functions of two variables assumes the function is differentiable. Do the above assumptions give us differentiability or is there a version of the Mean Value Theorem for some partial derivatives?
I think the point here is that you are working on a compact set(rectangle) and can keep $t$ fixed. I.e., you are considering a function $f_t(x)$ for a fixed $t$ on this rectangle and can then get Lipschitz of this function by mean-value-theorem. You will get that $|f_t(x_1)-f_t(x_2)| \leq L_{t,K}|x_1-x_2|$. Now your Lipschitz-constant (by the mean-value theorem for 1-dimension) is the partial-derivative w.r.t. x of your original function at some $(t,c), c \in [x_1,x_2]$. Now you can take the supremum on both sides of the inequality over $t$, i.e.
$ \underset{t}{sup}|f_t(x_1)-f_t(x_2)| \leq \|x_1-x_2| \underset{t}{sup} L_{t,K} = |x_1-x_2|\underset{t,x}{sup} |\frac{\partial f}{\partial x}(t,x)| $. Where the equality will follow from continuity (but thats only important if you really want to determine the Lipschitz-constant, if not, you can also assume there is a second lesser/equal sign). As your partial derivative is continuos and your set is compact, the supremum on the right side will be finite.