Let $\{f_\alpha: \alpha\in A\}$ be a set of continuous functions $f_\alpha: X\to Y$. If $A, X, Y$ are all topological spaces, then is is true that $F: X\times A\to Y$ given by $F(x, a) = f_a(x)$ is continuous?
I'm also interested in a more specific application of this statement. In my undergraduate topology class, the definition of a space $X$ being path-connected is that for any $x_1, x_2\in X$ there exists a continuous $h_{x_1, x_2}: [0, 1]\to X$. If we define $H: X\times X\times [0, 1]\to X$ by $H(x_1, x_2, t) = h_{x_1, x_2}(t)$, then is $H$ necessarily continuous?
This is not true. Consider the following family $f_t:\mathbb R\to\mathbb R$ for $t\in[0,1]$ defined by $$f_t(x)=\left\{\begin{array}{lcl} x&:&0\leq t\leq\frac{1}{2}\\ -x&:&\frac{1}{2}<x<1. \end{array}\right.$$ Then $f_t$ is continuous for all $t$, but the map $(x,t)\mapsto f_t(x)$ is not continuous on $\mathbb R\times[0,1]$.
The problem with the case you are concerned with is that path connectedness ensures the existence of a path, not that the path depend continuously on the points. So you're question is not well-defined. There are many choices for the path, hence many choices for the function $H$. and not all of them will be continuous. For example, we can take $H:\mathbb R^2\times \mathbb R^2\times[0,1]\to\mathbb R^2$ to be $$H(x_1,x_2,t)=(1-t)x_1+tx_2$$ for $x_1\neq(1,0)$ and ,$x_2\neq(-1,0)$, and take $$H((1,0),(-1,0),t)=(\cos(\pi t),\sin(\pi t)),$$ and obtain a non-continuous map. That's not to say you can't define a map $H:X\times X\times[0,1]\to X$ which will be continuous with $H(x_1,x_2,0)=x_1$ and $H(x_1,x_2,1)=x_2$ for all $x_1,x_2\in X$, but as the example suggests, it's certainly not guaranteed.