Uniform Limit of Riemannian Isometries is also a Riemannian Isometry This is problem 6-3 in Lee's Riemannian Geomtry (I have changed some notation below).

Problem. Let $M$ and $N$ be Riemannian manifolds and let $f_n:M\to N$ be a sequence of Riemannian isometries that converge uniformly to a map $f:M\to N$. Then show that $f$ too is a Riemannian isometry.

Here is what I have thought. We use the face that a map $\varphi:M\to N$ is a Riemannian isometry if and only if $\varphi$ is an isometry under the metrics (in the sense of metric spaces) induced on $M$ and $N$ by the a respective Riemannian metrics.
Thus it is sufficient to prove that $f$ is an isometry in the sense of metric spaces, that is, a metric isometry.
Using the uniform convergence and isometry property of $f_n$'s, it is easy to argue that $f$ is injective and continuous.
I am unable to show that $f$ is surjective. I tried the following: Let $q$ be an arbitrary point in $N$ and let $p_n\in M$ be the unique point such that $f_n(p_n)=q$. It follows that $(p_n)$ is a Cauchy sequence in $M$. But since $M$ is not necessarily complete, we cannot infer that $(p_n)$ converges. If it does converge, say to a point $p$, then once can argue that $f(p)=q$, but there is no guarantee for its convergence.
EDIT: To show that $f$ is surjective, we follow Jack Lee's hint below. Let $y\in N$ be a point in the closure of the image of $f$. Let $U$ be a uniformly normal neighborhood of $y$ in $N$, and let $x\in M$ be a point such that $f(x)\in U$. Let $u\in T_{f(x)}N$ be such that $\exp_{f(x)}u=y$, and that the geodesic segment $t\mapsto \exp_{f(x)}(tu):[0, 1]\to N$ is contained in $U$. For each $n$, let $x_n$ be the unique point in $M$ such that $f_n(x_n)=x$. Since each $f_n$ is an isometry, and since isometries pullback geodesics to geodesics, for each $n$ we can find $v_n\in T_{x_n}M$ such that $f_n(\exp_{x_n}v_n)=y$. Expplicitly, $v_n$ is the unique vector in $T_{x_n}M$ such that $df_n|_{x_n}(v_n)=u$. Thus $|v_n|=|u|$. But $|u|$ is same as the length of the aforementioned geodesic segment joining $f(x)$ to $y$. This in turn is same as the distance between $f(x)$ and $y$, for $U$ was chosen to be uniformly normal. Therefore the length of the geodesic segment $t\mapsto \exp_{x_n}(tv_n):[0, 1]\to M$ is same as $d(f(x), y)$ as is also equal to $|v_n|$. What we have just shown is that $|v_n|$ is constant an is equal to $d(f(x), y)$ for each $n$.
It is easy to see that $x_n$ converges to $x$. So for $n$ large enough, we have each $x_n$ in a geodesic ball $B$ about $x$. Without loss of generality we assume that $x_n\in B$ for all $n$. Let $w_n$ be the vector in $T_xM$ obtained by parallel transporting $v_n$ to $x$ via the unique geodesic path joining $x_n$ and $x$ in $B$. Since $w_n$ has the same magnitude as $v_n$, and since the magnitude of $v_n$'s is constant, we have that some subsequence of $(w_n)$ converges to a vector $w\in T_xM$. Again, we may assume that $w_n$ itself converges to $w$, passing to a subsequence if necessary. Now let $V$ be a neighborhood of $x$ in $M$ such that the geodesic exponential map is defined in some open set of $TV$ which contains $V$. We may assume that each $x_n\in V$.
Now the idea is this. Consider the map $\phi$ which eats a pair $(\alpha_n, \beta_n)\in T_xM\times T_xM$, and spits out a point in $TM$ whose fibre is that of $\exp_x(\beta_n)$, and tangent vector is the parallel transport of $\beta_n$ along the geodesic segment $t\mapsto \exp_x(t\alpha_n)$. Of course, this map in only defined on an open subset of $T_xM\times T_xM$. By this post, this is a smooth map. Compose this map by the exponential map. Thus this composite takes $(\alpha_n, w_n)$, where $\alpha_n$ is the "direction which aims at $x_n$", and produces $\exp_{x_n}(v_n)$.
So we have $\phi(\alpha_n, w_n)= \exp_{x_n}(v_n)$, giving $f_n(\phi(\alpha_n, w_n))= f_n(\exp_{x_n}(V_n))= y$. Using uniform convergence of $f_n$ to $f$, we have
$$\lim_{n\to \infty} f_n(\phi(\alpha_n, w_n))= \lim_{n\to infty}f(\phi(\alpha_n, w_n))$$
Note that $\lim_{n\to\infty}{\alpha_n}=0$, and thus we have
$$y=\lim_{n\to\infty}f_n(\exp_{x_n}(v_n)) = \lim_{n\to \infty} f_n(\phi(\alpha_n, w_n))= f(\phi(0, w)) = f(\exp_xw)$$
showing that $f$ is surjctive.
 A: I'm glad you asked this question. It has occurred to me occasionally that this problem is probably too hard without some kind of hint, so I finally sat down and tried to work out a reasonable hint. 
Try this: To show that $f$ is surjective, suppose $y$ is a limit point of $f(M)$. Choose $x\in M$ such that $f(x)$ lies in a uniformly normal neighborhood of $y$, and show there exists a convergent sequence of points $(x_n,v_n)\in TM$ such that $f_n(x_n) = f(x)$
and $f_n \big(\exp_{x_n} v_n\big)=y$.
Let me know if this helps.
A: If $f_n : (M,g_n)\rightarrow (N,g)$ is an isometry between Riemannian manifolds (that may be compact manifolds, i.e., not complete), then assume that $f_n\rightarrow f$. 
Here Riemannian metric $g_n$ is induced from $g$.
Not surjective case : Consider $M$ is $[0,1]^2$ and $N$ is $\mathbb{R}^3$. If $f_n(M)$ are "folding", then $f(M)$ may have image $[0,1]\times [0,\frac{1}{2}]$. Hence $f$ is not one-to-one.
Sujective case : Note that it is bijective so that we have a claim that uniform limit $f$ of isometries on $N$ is isometric. Since $$ d_g(x,y)= d_g(f_n(x),f_n(y)) \rightarrow d_g (f(x),f(y)) $$ then $f$ is isometric.
