Close maps are homotopic

Statement
Suppose $$X$$ is a topological space and $$Y$$ is a compact subset in Euclidean space, such that there exists an open subset $$U \supset Y$$ such that $$Y$$ is a strong deformation retract of $$U$$. Then, there exists $$\epsilon > 0$$ such that if two maps $$f_1,f_2:X \to Y$$ are $$\epsilon$$-close, in the sense:

$$d(f_1(x), f_2(x)) < \epsilon$$

(where $$d$$ denotes the Euclidean distance) then $$f_1$$ and $$f_2$$ are homotopic.

Alternative interpretations
A concrete interpretation of this is as follows. Suppose we view $$Y$$ as a compact metric space with the metric induced as a subset of $$\R^n$$. Then we can give $$C(X,Y)$$ the topology of uniform convergence. There is a natural map:

$$C(X,Y) \to [X,Y]$$

where $$[X,Y]$$ denotes the space of homotopy classes of continuous maps from $$X$$ to $$Y$$. The above result says that the above map is continuous if we give $$[X,Y]$$ the discrete topology. This interpretation follows because for every function in a homotopy class, the $$\epsilon$$-neighbourhood of that function is also in the same homotopy class.

The advantage of this interpretation is that for $$Y$$ a compact metric space, the topology of uniform convergence coincides with the compact-open topology, which can be defined without reference to the explicit metric. Thus, we can state the result more abstractly as:

If $$X$$ is a topological space and $$Y$$ is a compact metrizable space, give $$C(X,Y)$$ the compact-open topology and $$[X,Y]$$ the discrete topology. Then the mapping:

$$C(X,Y) \to [X,Y]$$

is continuous.

Converse
A converse to this statement exists, but under different hypotheses; we need to assume that the space $$X$$ is compact and $$Y$$ just needs to be a metric space.