Continuous Functions on Compact Space form Banach Space

Theorem
Let $X$ be a compact Hausdorff space.

Let $Y$ be a Banach space.

Let $\mathcal C = \mathcal C(X;Y)$ be the set of all continuous mappings $X \to Y$.

Let $\|\cdot\|_{\infty}$ be the supremum norm on $\mathcal C$.

Then $\left({\mathcal C, \left\|{\cdot}\right\|}\right)$ is a Banach space.

Proof
We have that the set of continuous mappings $X \to Y$ is a subset of the set $Y^X$ of all mappings $X \to Y$.

Therefore by Vector Space of All Mappings, we need only show that $\mathcal C$ is a subspace of $Y^X$.

By the One-Step Vector Subspace Test we need only show that $\mathcal C$ is closed under linear combinations (clearly $\mathcal C$ contains $0$).

But this is shown by the Combined Sum Rule for Continuous Functions.

We have Supremum Norm is Norm.

It remains to be shown that $\mathcal C$ is complete.

But this is precisely the statement of Uniform Limit of Continuous Functions is Continuous.