How to prove that $\bigcup S_n = (0,1.5)$ if $S_n=\{\frac{1}{n}\le x< 1+\frac{1}{n}\}$? 
Given that $S_n=\{x\in \mathbb R|\frac{1}{n}\le x< 1+\frac{1}{n}\}$, $n\in \mathbb N - \{0,1\}$ I need to find $\bigcup S_n$.

If we plug in $n=2$ then $0.5 \le x < 1.5$.
If $n\to \infty$ then $0<x\le 1$. Thefore $\bigcup S_n=(0,1.5)$.
I'm having trouble proving this.
Direction 1: $\bigcup S_n \subseteq (0,1.5)$:
If some $x \in S_n$ then there exists some $m$ such that $\frac{1}{m}\le x<1+\frac{1}{m}$ then because it's a set of unions then $S_n \in \bigcup S_n$.
Direction 2: $(0,1.5)\subseteq \bigcup S_n$:
By Archimedean principle for $x>0$ exists $n\in \mathbb N$ such that $0<\frac{1}{n}<x$. 
So if we choose some $n \in \mathbb N-\{0,1\}$ then $0< \frac{1}{n} < 1$ then $(0,1.5)\subseteq \bigcup S_n$.
I'm really not sure about direction 2 and even if I'm right I'd appreciate an explanation of the 2nd direction because I don't really understand it.
 A: As comment, for $\;n=2\;$ we already have $\;\left[\frac12,\,\frac32\right)\subset\bigcup S_n\;$ . Now, observe that
$$\left[\frac1n,\,1+\frac1n\right)\supset\left[\frac1{n+1},\,1+\frac1{n+1}\right)$$
so to get the whole union pass to limits:
$$\begin{cases}\frac1n\xrightarrow[n\to\infty]{}0\\{}\\
1+\frac1n\xrightarrow[n\to\infty]{}1\end{cases}\;\;\;\implies\bigcup_{2\le n\in\Bbb N}S_n=\left(0,\,\frac32\right)$$
If the above doesn't make it for you, we can argue as follows:
$$x\in\left(0,\,\frac32\right)\implies 0<x <\frac32\implies\exists\,n\in\Bbb N\;\;s.t.\;\;\frac1n<x<1+\frac1n\implies $$
$$\implies x\in\left(\frac1n,\,1+\frac1n\right)\subset \bigcup S_n$$
A: Let's try;
$x \gt 0.$
$(0,1.5) \subset \cup S_n.$
1) Let $0 <x \le 0.5:$
Consider $1/x$ , real.
Archimedes :
There is a $n_0 \in \mathbb{Z^+}$ such that
$n_0 \gt 1/x$ , or
$x>1/n_0.$ 
Combining: $1/n_0 < x <1+1/n_0$, hence
$x \in S_{n_0}.$
2) Let $0.5 < x <1.5$.
Then choose $n_1= 2$, and we have
$1/n_1 <x <1+1/n_1$, 
hence $x \in S_{n_1}$.
Altogether: $(0,1.5) \subset \cup S_n$.
A: For what it is worth, I was wondering if there was a way to calculate the 1.5 instead of guessing it first, and I found the following proof.  This may or may not be to your taste; see e.g. EWD1300 if you are interested in the background of this style of proof and notation.
In this answer $\;x\;$ is real and $\;n\;$ is integer.$%
\require{begingroup}
\begingroup
\newcommand{\calc}{\begin{align} \quad &}
\newcommand{\op}[1]{\\ #1 \quad & \quad \unicode{x201c}}
\newcommand{\hints}[1]{\mbox{#1} \\ \quad & \quad \phantom{\unicode{x201c}} }
\newcommand{\hint}[1]{\mbox{#1} \unicode{x201d} \\ \quad & }
\newcommand{\endcalc}{\end{align}}
\newcommand{\Ref}[1]{\text{(#1)}}
\newcommand{\true}{\text{true}}
\newcommand{\ceil}[1]{\left\lceil #1 \right\rceil}
%$

We calculate which $\;x\;$ are in $\;\bigcup_{n \ge 2}S_n\;$:
$$\calc
    x \in \bigcup_{n \ge 2}\left\{x \mid \tfrac 1 n \le x \lt 1 + \tfrac 1 n \right\}
\op\equiv\hint{expand definition of $\;\bigcup\;$}
    \langle \exists n :: n \ge 2 \;\land\; \tfrac 1 n \le x \lt 1 + \tfrac 1 n \rangle
\op\equiv\hint{multiply by $\;n\;$ -- work towards isolating $\;n\;$}
    \langle \exists n :: n \ge 2 \;\land\; 1 \le nx \;\land\; n(x-1) \lt 1 \rangle
\op\equiv\hints{divide middle inequality by $\;x\;$, which must be $\;{}>0\;$}\hint{-- further isolating $\;n\;$}
    x \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \ge \tfrac 1 x \;\land\; n(x-1) \lt 1 \rangle
    \tag{*}
\endcalc$$
Now, to isolate $\;n\;$ in the last inequality, we need to divide by $\;x-1\;$, which can be positive, negative, or zero, so a case split seems unavoidable.
Case $\;x<1\;$.
$$\calc
    \Ref{*}
\op\equiv\hint{divide by $\;x-1\;$, which is negative}
    x \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \ge \tfrac 1 x \;\land\; n \gt \tfrac 1 {x-1} \rangle
\op\equiv\hint{choose a large enough $\;n\;$}
    x \gt 0
\endcalc$$
Case $\;x=1\;$.
$$\calc
    \Ref{*}
\op\equiv\hint{substitute $\;x:=1\;$}
    1 \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \ge 1 \;\land\; 0 \lt 1 \rangle
\op\equiv\hint{simplify; choose a large enough $\;n\;$}
    \true
\endcalc$$
Case $\;x>1\;$.
$$\calc
    \Ref{*}
\op\equiv\hints{use $\;2 \gt \tfrac 1 x\;$ to remove $\;n \ge \tfrac 1 x\;$;}\hint{divide by $\;x-1\;$, which is positive}
    x \gt 0 \;\land\; \langle \exists n :: n \ge 2 \;\land\; n \lt \tfrac 1 {x-1} \rangle
\op\equiv\hints{apply $\Ref{0}$ below -- to make the rightmost}\hint{expression integer, for the next step}
    x \gt 0 \;\land\; \langle \exists n :: 2 \le n \lt \ceil{\tfrac 1 {x-1}} \rangle
\op{\equiv \tag{!}}\hint{arithmetic: simplify using total order on integers}
    x \gt 0 \;\land\; 2 \lt \ceil{\tfrac 1 {x-1}}
\op\equiv\hint{apply $\Ref{0}$ again -- to simplify}
    x \gt 0 \;\land\; 2 \lt \tfrac 1 {x-1}
\op\equiv\hint{arithmetic using $\;x>1\;$}
    0 \lt x \lt 1.5
\endcalc$$
Combining the above, we get
$$\calc
    \Ref{*}
\op\equiv\hint{combine the above cases}
    0 < x < 1 \;\lor\; x = 1 \;\lor\; 1 < x < 1.5
\op\equiv\hint{}
    x \in (0,1.5)
\endcalc$$
which completes the proof.

Here we used the 'ceiling' or 'round up' notation $\;\ceil\ldots\;$, and we used
$$
\tag{0}
m < y \;\equiv\; m < \ceil y
$$
for any integer $\;m\;$ and real $\;y\;$, to convert between integers and reals.  That is what allowed the key step $\Ref{!}$.
$%
\endgroup
%$
