You’ve misunderstood the proof that the set of Ulam numbers is infinite. It does not say that $u_n+u_{n-1}$ will always be an Ulam number: it says that if there are only finitely many Ulam numbers, and if $u_n$ and $u_{n-1}$ are the two largest Ulam numbers, then $u_n+u_{n-1}$ would have to be another Ulam number. This contradicts the assumption that $u_n$ and $u_{n-1}$ are the two largest Ulam numbers. But we know that given any finite set of numbers with at least two members, we can always pick out the two largest, so if the set of Ulam numbers were finite, we could certainly choose $u_n$ and $u_{n-1}$ to be the two largest Ulam numbers. Thus, the real problem is the assumption that there are only finitely many Ulam numbers, and the contradiction shows that this cannot be the case, i.e., that there must be infinitely many Ulam numbers.
To recapitulate the argument in a little more detail:
If $U$ is a finite set of positive integers, the largest possible sum of two members of $U$ is obtained by adding the two largest members of $U$, and every other pair will have a smaller sum. Thus, if $U$ were the entire set of Ulam numbers, and if $u_n$ and $u_{n-1}$ were the two largest members of $U$, then $u_n+u_{n-1}$ would be the largest possible sum of two members of $U$, and $u_n$ and $u_{n-1}$ would be the only two members of $U$ having that sum. But that would by definition make $u_n+u_{n-1}$ an Ulam number not in $U$, contradicting the supposition that $U$ contained every Ulam number. Thus, no such set $U$ can exist, and the set of Ulam numbers must be infinite.
One easy and insightful way is to use the proof below. It essentially constructs $\rm\:gcd\:$ from $\rm\:lcm\:$ by employing duality between minimal and maximal elements - see the Remark below. This is essentially how the linked Wikipedia proof works, but there the innate duality is obfuscated by the presentation. Below is a proof structured so that this fundamental duality is brought to the fore.
$\rm{\bf Theorem}\quad c\mid a,b\iff c\mid d,\ \ $ for $\rm\ \ d = ab/lcm(a,b).\ $ $\rm\color{#0a0}{Hence}$ $\rm\ d = gcd(a,b)$
$\rm{\bf Proof}\qquad\ \ \, c\mid a,b \iff a,b\mid ab/c \iff lcm(a,b)\mid ab/c \iff c\mid ab/lcm(a,b)$
$\rm\color{#0a0}{Generally}\,$ if $\rm\, c\mid a,b\iff c\mid d\ $ then $\rm\ d = \gcd(a,b)\ $ up to unit factors, i.e. they're associate.
Indeed setting $\rm\:c = d\:$ in direction $(\Leftarrow)$ shows that $\rm\:d\mid a,b,\:$ i.e. $\rm\:d\:$ is a common divisor of $\rm\:a,b.\:$ Conversely, by direction $(\Rightarrow)$ we deduce that $\rm\:d\:$ is divisible by every common divisor $\rm\:c\:$ of $\rm\:a,b,\:$ thus $\rm\:c\mid d\:\Rightarrow\: c\le d,\:$ so $\rm\:d\:$ is a greatest common divisor (both divisibility and magnitude-wise).
Remark $\ $ The proof shows that, in any domain, if $\rm\:lcm(a,b)\:$ exists then $\rm\:gcd(a,b)\:$ exists and $\rm\ gcd(a,b)\,lcm(a,b) = ab\ $ up to unit factors, i.e. they are associate. The innate duality in the proof is clarified by employing the involution $\rm\ x'\! = ab/x\ $ on the divisors of $\rm\:ab.\:$ Let's rewrite the proof using this involution (reflection).
Notice that $\rm\ x\,\mid\, y\:\color{#c00}\iff\: y'\mid x'\,\ $ by $\smash[t]{\,\ \rm\dfrac{y}x = \dfrac{x'}{y'} \ }$ by $\rm\, \ yy' = ab = xx',\ $ so rewriting using this
$\begin{eqnarray}\rm the\ proof\ \ \ c\mid a,b &\iff&\rm b,\,a\mid ab/c &\iff&\rm lcm(b,\,a)\mid ab/c &\iff&\rm c\mid ab/lcm(b,a)\\[.5em]
\rm becomes\ \ \ \ c\mid a,b &\color{#c00}\iff&\rm a',b'\mid c' &\iff&\rm lcm(a',b')\mid c' &\color{#c00}\iff&\rm c\mid lcm(a',b')'\end{eqnarray}$
Now the innate duality is clear: $\rm\ gcd(a,b)\,=\,lcm(a',b')'\ $ by the $\rm\color{#0a0}{above}$ gcd characterization.
Best Answer
Your proof is fine. Another way to say the same thing is: $x\mathbin|n$ and $x\mathbin|n+1$ would imply that $x\mathbin|(n+1)-n=1$, contradicting with the premise that $x>1$.