Suppose $\{ f_n \}$ converges uniformly on $(-1,1)$. Suppose also that $f_n(-1)$ and $f_n(1)$ converge. Then $\{ f_n \}$ converges uniformly on $\lbrack -1, 1 \rbrack$.
Attempt:
Suppose $\{ f_n \}$ does not converge uniformly to $f(x)$ on the closed interval. Then there is a sequence $x_n$ in $\lbrack -1 , 1 \rbrack$ such that for some $\epsilon_0 > 0$, $$ |f_n(x_n) - f(x)| \geq \epsilon_0. $$ Note that $x_n \neq \pm 1$ for all $n$. If it did, then $$ |f_n(\pm1) - f(x)| \geq \epsilon_0, $$ contradicting convergence. So $x_n \in (-1,1)$ which contradicts that $f_n$ converges uniformly on $(-1,1).$
I think you're over complicating matters.
Let $\varepsilon>0$. By uniform convergence, there is an $N$ such that $|f_n(x)-f(x)|<\varepsilon$ for every $x\in (-1,1)$ and every $n\ge N$. Also, since $f_n(1)\to f(1)$, there is some $N'$ such that $|f_n(1)-f(1)|<\varepsilon$ for every $n\ge N'$. Similarly, there is some $N''$ such that $|f_n(-1)-f(-1)|<\varepsilon$ for every $n\ge N''$. Now, if $n\ge \max\{N,N',N''\}$, then $|f_n(x)-f(x)|<\varepsilon$ for every $x\in [-1,1]$.