Is the uniform limit of (bijective) metric space isometries also a (bijective) metric space isometry?

Suppose $$X, Y$$ are metric spaces and $$\varphi_{i}:X\rightarrow Y$$ are bijective isometries that uniformly converge to a map $$\varphi:X\rightarrow Y$$. Is $$\varphi$$ a bijective isometry?

According to this post and the answer. There is an analogous result in Riemannian geometry if you assume $$X, Y$$ are connected Riemannian manifolds and $$d_{X}, d_{Y}$$ are the distance functions. However, the hint by Jack Lee to show that $$\varphi$$ is surjective relies on Riemannian geometry.

Does the result that $$\varphi$$ is surjective hold in general for metric spaces? Or is this result specific to Riemannian manifolds? Is there a counterexample?

Do we need some assumption about Cauchy completeness?

• For every $\varepsilon>0,$ let $N$ so large that $||\varphi_i-\varphi_j||\leq \varepsilon$ for $i,j\geq N$. Now let $y\in Y$ and $x_{\varepsilon}=\varphi_N^{-1}(y)$. Then, $d_Y(\varphi(x_{\varepsilon}),\varphi_{N}(x))\leq \varepsilon$. Then, necessarily, $\lim_{n\to\infty}\varphi(x_{1/n})=y,$ so the question reduces to whether or not the sequence $x_{1/n}$ converges (or has a convergent subsequence). I think this becomes a question of whether $X$ is locally compact (the $x_{1/n}$ sequence must be bounded). Jul 17 '19 at 6:47

Let $$\{\varphi_n:X\rightarrow Y\}$$ be the sequence of uniformly converging bijective isometries and $$\varphi:X\rightarrow Y$$ its limit. It is clear, that $$\varphi$$ is an injective isometry, as uniform convergence implies punctual convergence. So we only have to show that $$\varphi$$ is surjective.
Without loss of generality, by passing to a subsequence if necessary, assume that $$d_\infty(\varphi_n,\varphi)<1/n$$ where $$d_\infty(\psi,\tilde{\psi}):=\sup_{x\in X}d(\psi(x),\tilde{\psi}(x))$$.
Let $$y\in Y$$. We will construct $$x\in X$$ such that $$\varphi(x)=y$$. This will show surjectivity. Let the sequence $$\{x_n\}$$ be given by $$x_n:=\varphi_n^{-1}(y).$$ Note that $$d(\varphi(x_n),y)=d(\varphi(x_n),\varphi_n(x_n)) and so $$\lim_{n\to\infty}\varphi(x_n)=y.$$ Hence it is enough to show that $$\{x_n\}$$ is convergent. Until here I only repeated what @WoolierThanThou said.
Now, for all $$k,l\geq 2n$$ and $$m$$, $$d(x_k,x_l)=d(\varphi_m(x_k),\varphi_m(x_l))\leq d(\varphi_m(x_k),y)+d(y,\varphi_m(x_l))$$ where the first equality follows from the fact that $$\varphi_m$$ is an isometry and the second inequality is the triangular inequality. Taking the limit $$m\to\infty$$, we get that for all $$k,l\geq 2n$$, $$d(x_k,x_l)\leq d(\varphi(x_k),y)+d(y,\varphi(x_l))< 1/k+1/l\leq 1/(2n)+1/(2n)\leq 1/n,$$ where the second inequality follows from the inequality in the previous paragraph. Hence $$\{x_n\}$$ is a Cauchy sequence and thus convergent by the initial assumption. We are done.