Q:

Conintuity at a point where two continuous functions meet

I'm aware the pasting lemma, but in this course I do not have that tool.  I was wondering if I could show continuity with sequences.
So here is my problem.  We have two continuous functions $f:[0,1] \to X$ and $g:[1,2]\to X$ such that $f(1)=g(1)$.  I want to show that it is continuous at $x=1$.
So let $x_n$ be a sequence such that $x_n=1-1/n$ and and $x'_n=1+1/n$.  Then $\lim f(x_n)=f(1)$ and $\lim g(x'_n)=g(1)$ since both are continuous.  So its continuous at the point $x=1$

A:

You did not prove that the "pasted" function (call it, say, $h$) is continuous. You have to prove that $f(x_n)$ converges for any $x_n$. Here you proved it for two particular sequences.
If you want to use sequences, your proof must start with "Let $x_n$ be a sequence in $[0,2]$ converging to 1."
For instance, you may split this sequence in two subsequences, one of which lies in $[0,1]$, the other in $[1,2]$.
EDIT : Subsequences may not be the best way to go (it works but actually writing it is kind of boring). 
This is what i suggest.
Let $x_n$ be a sequence in $[0,2]$ converging to 1. We create two sequences, $y_n = \min(x_n,1)$ and $z_n = \max(x_n,1)$. Actually, what we're doing is forcing $y_n$ to live in [0,1] and $z_n$ to live in [1,2] by replacing undesired elements of $x_n$ by 1's.
So $(y_n) \to 1$ in $[0,1]$ and $(z_n) \to 1 $ in $[1,2]$. 
Thus $h(y_n)=f(y_n) \to f(1)=h(1)$ and $h(y_n)=g(z_n) \to g(1)=h(1)$.
Now let $\epsilon >0$. There is $M$ s.t. for $n>M$, $|h(y_n)-h(1)| < \epsilon$. There is $N$ s.t. for $n>N$, $|h(z_n)-h(1)| < \epsilon$. 
So for $n>\max(N,M)$, either $x_n=y_n$ or $x_n=z_n$, in both cases, $|h(x_n)-h(1)| < \epsilon$. 

