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The cumulative distribution function of the uniform distribution on $[0,1/n]$ isThe constant zero has cumulative distribution function $F(t)=1_{[0,\infty )}(t)$. Its continuity points are precisely $t\ne 0$. For fixed $t<0$, $F_n(t)=0=F(t)$; for fixed $t>0$, $F_n(t)=1=F(t)$ eventually. This proves convergence in distribution to zero directly from the definition.

Also, $\mathbb{P}(0\le U_n\le 1/n)=1$. For every $\epsilon >0$ and $n>1/\epsilon$, $\mathbb{P}(|U_n|>\epsilon )=0$. HenceThe convergence in probability conclusion holds for any realization of these random variables on a common probability space; no independence of random variables is needed.

The definition of convergence in distribution gives $F_n(t)\to F(t)$ at each continuity point $t$ of $F$. We must handle a moving argument rather than assume this convergence is uniform.

A cumulative distribution function has at most countably many discontinuities: each is a jump of positive size, and there are at most $r$ jumps of size at least $1/r$, since their sizes sum to at most one. Thus we can choose continuity points $u_j<c<v_j$ with $u_j\uparrow c$ and $v_j\downarrow c$. For fixed $j$, eventually $u_j<c_n<v_j$. The monotonicity of each cumulative distribution function impliesFirst let $n\to \infty$, using the definition of convergence in distribution at $u_j,v_j$. Then let $j\to \infty$, using continuity at $c$:No result relating different convergence of random variables has been assumed.