Let $f_n$ a sequence of contractions on a metric space $(Y,d)$, with a Lipschitz constant $0<\lambda<1$. Suppose that for all $y\in Y$ the sequence $f_n(y)$ converges to $f(y)$. Then $F$ is also a contraction with Lipschitz constant $\lambda$.
My proof is as follows: \begin{align*} d(f(y_1),f(y_2)) \leq d(f(y_1),f_n(y_1)) + d(f_n(y_1),f_n(y_2)) + d(f_n(y_2),f(y_2)) \leq d(f(y_1),f_n(y_1)) +\lambda\cdot d(y_1,y_2) + d(f_n(y_2),f(y_2)) \end{align*} Then taking $n\to\infty$ I get $d(f(y_1),f(y_2))\leq \lambda\cdot d(y_1,y_2)$, so $f$ is a contraction with Lipschitz constant $\lambda$.
Is the proof correct? Or do I need $f_n\to f$ uniformly?