My professor provided a different, and less illuminating proof; but I wrote this one and would like to know if it's legitimate.
Let $f:(0, 1] \rightarrow \mathbb{R}~$ be uniformly continuous. $f$ is bounded on this domain. Thus choosing a sequence $\{x_n\} \subset [0, 1]$ such that for all $n, ~ x_{n+1} < x_n$, and $\{x_n\} \rightarrow 0$, $\{f(x_n)\}$ is also a bounded sequence in $\mathbb{R}$ and thus contains a convergent subsequence $\{x_{n_{k}}\}$. Call its limit $L_1$.
Fix $\epsilon/2 >0$. There exists $N$ such that for any ${n_k}>N, ~|f(x_{n_k}) - L_1|<\epsilon/2$.
For the same $\epsilon$, the uniform continuity of $f$ furnishes a $\delta$ such that $|x-y|< \delta$ implies that $|f(x) - f(y)|<\epsilon/2$.
Choose $M$ sufficiently large that for all $n, n_{k} < N, ~ |x_{n}-x_{n_k}|<\delta$ (this value is guaranteed to exist because all subsequences of $x_n$ approach arbitrarily close to $0$. Fix $P = \max(N, M)$.
Then, for all $n, n_k>P$
$|f(x_n) - L| \leq |f(x_n) - f(x_{n_k})| + |f(x_{n_k})-L_1| <\epsilon$.
edits as per suggestion
Choose another such sequence $\{x'_n\}$. All of the above statements hold true. For this sequence, $f(x'_n) \rightarrow L_2$
Now consider the sequence $X = \{x_1, x'_1, x_2, x_2'...\}$. This sequence also converges to $0$, and as such $f(X)$ is convergent, and has two convergent subsequences, which must converge to the same value: thus $L_1 = L_2$.
Thus the $\lim\limits_{x \rightarrow 0}f(x)$ exists.
End of edits
The other direction I prove by defining an extension onto the compact domain $[0,1]$; my professor did it the same way so I'm less concerned.
For this one, what would be most helpful would be feedback on flow and notation if it's more or less correct, or suggestions about fixes if it isn't. Thanks.
This is not an answer as such, just an alternative approach.
Often $\epsilon-\delta$ proofs obfuscate the salient points.
Note that if $f$ is uniformly continuous, then it maps Cauchy sequences into Cauchy sequences. This is the crucial element.
Pick a specific sequence that converges to $0$, say $x_n = {1 \over n}$, then since $f(x_n)$ is Cauchy we have $f(x_n) \to L$ for some $L$. This is our candidate value for the limit.
Now suppose $x_n' \to 0$ for some other sequence. As above we have $f(x_n') \to L'$ and we want to show that $L'=L$.
Now a little trick: Pick the sequence $x_1,x_1',....,x_n,x_n',...$. Since the sequence converges to $0$ it is Cauchy and hence $f(x_1),f(x_1'),....,f(x_n),f(x_n'),...$ is Cauchy and hence converges. Since there are subsequences that converge to $L,L'$ we see that $L'=L$.