Let $\mathfrak m$ be the maximal ideal of the ring of integers (valuation ring) of $K$, so your $U^{(1)}$ is $1 + \mathfrak m$. You want to prove $1+\mathfrak m$ looks like $\mathbf Z/p^a\mathbf Z \times \mathbf Z_p^d$ topologically for some $a \geq 0$ and $d = [K:\mathbf Q_p]$. You said you agree there is such an isomorphism "algebraically". What do you mean by that: an isomorphism as what kind of algebraic structures? It's not just as groups.
The key point is to be thinking about the isomorphism $1+\mathfrak m \to \mathbf Z/p^a\mathbf Z \times \mathbf Z_p^d$ by viewing the multiplicative group $1 + \mathfrak m$, which is a $\mathbf Z$-module by exponentiation, as a $\mathbf Z_p$-module by exponentiation: for each $u \in 1 + \mathfrak m$, the powers $u^n$ for $n \in \mathbf Z$ are $p$-adically uniformly continuous in $n$, so we can extend ordinary integral powers of $u$ to $p$-adic integer powers $u^x$ with $x \in \mathbf Z_p$ by $p$-adic continuity, and the effect of $\mathbf Z_p$-exponents on elements of $1+\mathfrak m$ is how $1+\mathfrak m$ turns into a $\mathbf Z_p$-module.
A torsion element $u$ of $1+\mathfrak m$ as a $\mathbf Z_p$-module is a $p$-power root of unity: if $u^x = 1$ where $x \not= 0$ then writing
$x = p^nv$ with $n \geq 0$ and $v \in \mathbf Z_p^\times$, we get
$1 = u^{p^nv} = (u^{p^n})^v$, so (raising both sides to the $1/v$-power) we get $1 = u^{p^n}$. Conversely, a $p$-power root of unity in $K$ must lie in $1 + \mathfrak m$ (basically because the only $p$-power roots of unity in a finite field is 1), so the torsion submodule of $1+\mathfrak m$ (as a $\mathbf Z_p$-module) is the $p$-power roots of unity in $K$. That is a finite group because a finite extension of $\mathbf Q_p$ has only finitely many $p$-power roots of unity in it: a root of unity of order $p^r$ has degree $p^{r-1}(p-1)$ over $\mathbf Q_p$ and that exceeds $[K:\mathbf Q_p]$ for large enough $r$. Finite subgroups of $K^\times$ are cyclic, so the torsion submodule $T$ of $1+\mathfrak m$ is a finite cyclic $p$-group.
Set $|T| = p^a$, so $T \cong \mathbf Z/p^a\mathbf Z$ as $\mathbf Z_p$-modules
($\mathbf Z_p$ acting by exponents on the left and by multiplication after reducing $p$-adic integers modulo $p^a$ on the right).
Provided $1 + \mathfrak m$ is a finitely generated $\mathbf Z_p$-module, the classification of finitely generated modules over a PID (such as the PID $\mathbf Z_p$) implies $1 + \mathfrak m \cong T \times \mathbf Z_p^d \cong \mathbf Z/p^a\mathbf Z \times \mathbf Z_p^d$ as $\mathbf Z_p$-modules for some $d\geq 0$. Of course $d \geq 1$ since $1+\mathfrak m$ is not finite. You said in your post that you already agree $1 + \mathfrak m \cong \mathbf Z/p^a\mathbf Z \times \mathbf Z_p^d$ algebraically, where $d = [K:\mathbf Q_p]$. I don't know if you meant that isomorphism exists as abelian groups or as $\mathbf Z_p$-modules. If you grant there is such an isomorphism as $\mathbf Z_p$-modules, then there is a root of unity $\zeta$ of order $p^a$ in $1+\mathfrak m$ and $d$ units
$u_1, \ldots, u_d$ in $1 + \mathfrak m$ that are $\mathbf Z_p$-independent (if $u_1^{x_1} \cdots u_d^{x_d} = 1$ then each $p$-adic integer $x_i$ is $0$) such that $1+\mathfrak m = \langle \zeta\rangle \times u_1^{\mathbf Z_p} \times \cdots \times u_d^{\mathbf Z_p}$. Then the mapping
$$
f \colon
\mathbf Z/p^a\mathbf Z \times \mathbf Z_p^d \to 1 + \mathfrak m
$$
given by $f(k,x_1,\ldots,x_d) = \zeta^ku_1^{x_1} \cdots u_d^{x_d}$ is a $\mathbf Z_p$-module isomorphism. This function $f$ is continuous using the product topology on its domain and the subspace topology on its codomain ($1+\mathfrak m$ as a subset of $K$) because (i) $u^x$ is $p$-adically continuous in $x$ for $u \in 1+\mathfrak m$ and $x \in \mathbf Z_p$ and (ii) multiplication in $1+\mathfrak m$ is continuous. The domain and codomain of $f$ are compact Hausdorff spaces, and a continuous bijection of compact Hausdorff spaces is a homeomorphism. Therefore if $f$ is an isomorphism of $\mathbf Z_p$-modules ("algebraically"), it is also a homeomorphism, which is what you wanted to understand.
