Can we use Lipschitz continuity to prove $f(x,y) = x^2+y^2$ is continuous for $x,y\in \mathbb{R}$? In Munkres' Topology, we have Lemma 21.4 which states that the addition, subtraction, and multiplication operations are continu­ous functions from $\mathbb{R} \times \mathbb{R}$ into $\mathbb{R}$. Can we use Lemma 21.4 and Lipschitz continuity to prove $f(x,y) = x^2+y^2$ is continuous for $x,y\in \mathbb{R}$?

No, we can't use Lipschitz continuity here because $f$ is _not_ Lipschitz due to the convexity of $f$.

According to the linked Wiki page, an equivalent definition of a Lipschitz function $f: X \to Y$ between two metric spaces $X,Y$ is that the difference quotient $$\forall\, x_1, x_2 \in X, \quad {\frac {d_{Y}(f(x_{1}),f(x_{2}))}{d_{X}(x_{1},x_{2})}}\leq K$$ is bouned by some constant $K$ (independent of $x_1,x_2$).

This is intuitively false as the graph of $f$ is the parabola $z = x^2$ rotated along the $z$-axis, and the slopes of a parabola are _unbounded_.

For an analytic argument, you may write

$$\frac{f(n+\frac1n,0) - f(n,0)}{\left(n+\frac1n\right) - n} = \dots = n \left( 2 + \frac{1}{n^2} \right) \xrightarrow[n \to +\infty]{} +\infty.$$

Therefore, the difference quotient is _not_ bounded, and $f$ is _not_ Lipschitz.