Consider a sequence of continuous mappings $f_n:M\to N$ for $n \in \mathbb{N}$, where $M$ and $N$ are compact smooth manifolds, or in general, compact and Hausdorff metric spaces.
Suppose that $f_n \to f$ in the $C^0$ topology, which in this case reduces simply to uniform convergence. Is it true that $f_n$ is homotopic to $f$ for every $n$ sufficiently large?
I believe that this is true because of the following: in our case, the property of two mappings $f$ and $g$ being homotopic is equivalent to being in the same path-connected component of the function space $C(M,N)$ in the $C^0$ topology, i.e., just the uniform convergence topology. And, if $f_n\to f$ in this topology, I believe it should be contained in the path connected component of $f$ for sufficiently large $n$ (but I don't know how to prove it).
My goal is to use this fact to argue that $deg(f_n) = deg(f)$ for all sufficiently large $n$.
For manifolds (or more generally absolute neighborhood retracts of euclidean spaces) consider embedding $N\subseteq \mathbb{R}^n$. Next pick a tubular neighborhood $U$ of $N$ in $\mathbb{R}^n$. There is a smooth retraction $r:U\rightarrow N$ and $U$ is an open subset of $\mathbb{R}^n$. For each $n\in \mathbb{N}$ consider the homotopy $h_n:[0,1]\times M\rightarrow \mathbb{R}^n$ given by formula $$h_n(t,m) = tf(m) + (1 - t)f_n(m)$$ Since $\{f_n\}_{n\in \mathbb{N}}$ converges uniformly to $f$, we derive that $$h_n([0,1]\times M)\subseteq U$$ for sufficiently large $n$. Then $H = r\cdot h_n:[0,1]\times M\rightarrow N$ is a homotopy between $f_n$ and $f$.
In the case of general compact metric spaces the result does not hold. For this pick $$M = pt,\,N = \{0\}\cup \bigg\{\frac{1}{n}\,|\,n\in \mathbb{N}\bigg\}$$ and consider continuous maps $f_n(pt) = \frac{1}{n}$ and $f(pt) = 0$. Then $f_n\rightarrow f$ uniformly, but $f_n$ are $f$ are never homotopic.