Since $X$ is not pseudocompact, there is a continuous function $f:X\to\Bbb R$ whose range is unbounded. Thus, we can choose a set $A=\{x_n:n\in\Bbb N\}\subseteq X$ such that $f(x_{n+1})>f(x_n)+1$ for each $n\in\Bbb N$. For $n\in\Bbb N$ let $V_n=f^{-1}\left[\left(f(x_n)-\frac12,f(x_n)+\frac12\right)\right]\;;$ then the family $\mathscr{V}=\{V_n:n\in\Bbb N\}$ is discrete. To see this, let $x\in X$ be arbitrary. Then $f(x)$ has an open nbhd $U$ in $\Bbb R$ that meets at most one of the open intervals $\left(x_n-\frac12,x_n+\frac12\right)$, and $f^{-1}[U]$ is an open nbhd of $x$ in $X$ that meets at most one member of $\mathscr{V}$. $X$ is Tikhonov, so for each $n\in\Bbb N$ there is a continuous function $g_n:X\to[0,1]$ such that $g_n(x_n)=1$, and $g_n(x)=0$ for all $x\in X\setminus V_n$.
Now let $h:A\to\Bbb R$ be arbitrary. Let $\hat h=\sum_{n\in\Bbb N}h(x_n)g_n:X\to\Bbb R$; clearly $\hat h\upharpoonright A=h$, and $\hat h$ is continuous because $\mathscr{V}$ is discrete. Thus, $A$ is $C$-embedded in $X$.