wrong proof of "locally lipschitz implies continuity"

Question

I think that I've proved that locally lipschitz implies continuity on metric space. But something must be wrong:

Let $(\mathfrak{X},d_1)$ and $(\mathfrak{Y},d_2)$ be metric spaces. If $\varphi :(\mathfrak{X},d_1)\longrightarrow (\mathfrak{Y},d_2)$ is locally lipschitz then $\varphi$ is continuous in $\mathfrak{X}$.

Suppose $x_0\in \mathfrak{X}$ and $\varepsilon>0$. As $\varphi$ is locally lipschitz there exists $\mu>0$ such that: $$\exists\lambda_{x_0}>0:d_2(\varphi(x),\varphi(y))\leq \lambda_{x_0}d_1(x,y) $$ for all $x,y\in B(x_0,\mu)$. So then for $\delta=\varepsilon/2\lambda$ we have that: $$d_1(\varphi(x),\varphi(x_0))\leq \frac{\varepsilon}{2}<\varepsilon$$ Many thanks !

Answer 1

As pointed out in the comments, you have to take $\delta\lt \varepsilon/(2\lambda_{x_0})$ and we should take $\delta<\mu$ in order to be sure that $d(x,x_0)\lt \mu$.

The result seems to be true: the notion of continuity is local, since we have to check continuity at each point. Local Lipschitzness gives more.