A uniform limit of continuous functions is continuous, so $C(X,Y)$ is closed in $Y^{X}$ under the uniform metric

Let $(X,\mathcal{T}_X)$ be a topological space (Topology on a set, open and closed sets, clopen sets, the closed-set axiomatisation, and the coarser/finer comparison) and let $(Y,d)$ be a metric space (Metric space: $d(x,y) = 0$ iff $x = y$, symmetry, and the triangle inequality; pseudometric and ultrametric) carrying its metric topology. Then:

1. The $\varepsilon/3$ criterion. Let $f : X \to Y$ be a function such that for every real $\varepsilon > 0$ there is a continuous $g : X \to Y$ with $d\big(f(x), g(x)\big) < \varepsilon \qquad \text{for every } x \in X .$ Then $f$ is continuous (Continuity of a map of topological spaces at a point and globally).
2. Uniform limit theorem. If $X$ is nonempty, $(f_k)$ is a sequence of continuous maps $X \to Y$ and $(f_k)$ converges uniformly to $f$ (Uniform convergence, and the topology of uniform convergence: the metric topology of the uniform metric on $Y^{X}$ and on $C(X,Y)$), then $f$ is continuous.
3. Closedness. If $X$ is nonempty, $C(X,Y)$ is a closed subset of $(Y^{X}, \bar\rho)$, the uniform metric being that of For a nonempty set $X$ and a metric space $(Y,d)$ the uniform metric $\bar\rho(f,g) = \sup_{x} \min\{d(f(x),g(x)), 1\}$ is a metric on $Y^{X}$.

The domain is an arbitrary topological space, not a metric space: nothing in the argument uses a distance in $X$. Only the target carries a metric, and it carries one because the hypothesis of claim 1 is a statement about distances in $Y$.

No choice principle is used, and claim 3 in particular is choice free. The proof of claim 3 instantiates one continuous $g$ for each $\varepsilon$ and uses it immediately, rather than manufacturing a sequence of them; a sequential argument through A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed would spend the Axiom of Countable Choice, and that route is deliberately not taken.

• The three thirds are the three legs of the estimate, and each is a different approximation: $f$ to $g$ at $x$, $g$ at $x$ to $g$ at $a$, and $g$ to $f$ at $a$. Only the middle one uses continuity, and only the outer two use that the approximation of $f$ by $g$ is uniform. If the approximation were merely pointwise, the third leg would still hold but the first would need an $\varepsilon$ depending on $x$, and the argument collapses; the companion page exhibits exactly that collapse.

• Claim 3 is what makes $C(X,Y)$ a complete space when $Y$ is complete, by the next item, and it is the reason the uniform topology and not the pointwise one is the natural home for limits of continuous functions. In the pointwise topology $C(X,Y)$ is in general not closed, and the companion page carries a witness on $[0,1]$.

• Why the choice-free route was taken. The usual proof of claim 3 shows that $C(X,Y)$ is sequentially closed and then invokes A point lies in the closure of $A$ iff some sequence in $A$ converges to it, and a set is closed iff it is sequentially closed to conclude closedness; that item's forward direction spends the Axiom of Countable Choice, since it manufactures a sequence out of adherence. The argument above instead works with the distance to the set directly and instantiates a single $g$ at each $\varepsilon$, so claim 3 is a theorem of ZF.