Yes, the conclusion holds. From reading the new comments, I guess my solution is something like a "supporting line".
Lemma: For $x_1 \in (0,c)$ and $x_2 \in (c,+\infty)$ from "$\phi$ convex at $c$" follows
$$\frac{\phi(x_2)-\phi(c)}{x_2-c} \ge \frac{\phi(c)-\phi(x_1)}{c-x_1}.$$
Geometrically that means the line through the points $(c,\phi(c))$ and $(x_2,\phi(x_2))$ has at least the slope of the line through $(x_1,\phi(x_1))$ and $(c,\phi(c))$.
Proof of Lemma: The choice of
$$\alpha=\frac{x_2-c}{x_2-x_1} \Rightarrow 1-\alpha=\frac{c-x_1}{x_2-x_1}$$
leads to $\alpha x_1+(1-\alpha)x_2=c$. Note that because of $x_1 < c < x_2$ the quotients are defined and we have $\alpha,1-\alpha \ge 0$, so $\alpha \in [0,1]$ as required. So we can apply "$\phi$ convex at $c$" and get
$$\phi(c) \le \frac{x_2-c}{x_2-x_1}\phi(x_1) + \frac{c-x_1}{x_2-x_1}\phi(x_2).$$
We want to isolate $\phi(x_2)$ and multiply the inequality with the positive $\frac{x_2-x_1}{c-x_1}$ to get
$$\frac{x_2-x_1}{c-x_1} \phi(c) \le \frac{x_2-c}{c-x_1}\phi(x_1) + \phi(x_2) \Rightarrow \phi(x_2) \ge \frac{x_2-x_1}{c-x_1} \phi(c) - \frac{x_2-c}{c-x_1}\phi(x_1).$$
From the latter inequality we subtract $\phi(c)$ on both sides and then devide by the positive $(x_2-c)$:
$$\frac{\phi(x_2)-\phi(c)}{x_2-c} \ge \frac{x_2-x_1}{(c-x_1)(x_2-c)}\phi(c) - \frac{1}{c-x_1}\phi(x_1) - \frac{\phi(c)}{x_2-c}.$$
The term on the right hand side can now be simplified by first uniting the coeffcients before $\phi(c)$ then cancelling $(x_2-c)$
$$\begin{eqnarray}
\frac{x_2-x_1}{(c-x_1)(x_2-c)}\phi(c) - \frac{1}{c-x_1}\phi(x_1) - \frac{\phi(c)}{x_2-c} & = & \frac{(x_2-x_1) - (c-x_1)}{(c-x_1)(x_2-c)}\phi(c) - \frac{1}{c-x_1}\phi(x_1)\\
& = & \frac{x_2-c}{(c-x_1)(x_2-c)}\phi(c) - \frac{1}{c-x_1}\phi(x_1)\\
& = & \frac{1}{c-x_1}\phi(c) - \frac{1}{c-x_1}\phi(x_1)\\
& = & \frac{\phi(c) - \phi(x_1)}{c-x_1},\\
\end{eqnarray}
$$
which finally proves the lemma!
Now for the OP's problem. Let the $x_i, \lambda_i$ be given for $i=1,2,\ldots,k$ and let's assume they fulfill the stated properties.
If all $x_i$ are at least $c$, then we have $x_i=c$ for all $i=1,2,\ldots,k$, then the conclusion is trivial. So assume at least one $x_i < c$ exists. W.l.o.g. assume that $x_1,x_2,\ldots x_r < c$ while $x_{r+1},\ldots,x_k \ge c$ for some $r \in \{1,2,\ldots,k\}$.
Let
$$m:=\max_{i=1,2,\ldots,r} \frac{\phi(c)-\phi(x_i)}{c-x_i}.$$
Let $i_{max}$ be one index where the maximum is realized.
I claim that all points $(x_i,\phi(x_i))$ lie on or above the line
$$y=m(x-c)+\phi(c)$$
passing through $(c,\phi(c)).$
If $i \le r$ we know by the definition of $m$ that
$$\frac{\phi(c)-\phi(x_i)}{c-x_i} \le m,$$
so multiplying with the positive $(c-x_i)$ yields
$$\phi(c)-\phi(x_i) \le m(c-x_i) \Rightarrow \phi(c) + m(x_i-c) \le \phi(x_i),$$
as desired.
If $x_i=c$, then $(x_i,\phi(x_i))=(c,\phi(c))$ obviously lies on the line.
If $x_i > c$, then we know by the lemma (remember that $x_i > c, x_{i_{max}} < c$, so the lemma can be applied) that
$$\frac{\phi(x_i)-\phi(c)}{x_i-c} \ge \frac{\phi(c)-\phi(x_{i_{max}})}{c-x_{i_{max}}} = m,$$
so we get immediately
$$\phi(x_i)-\phi(c) \ge m(x_i-c) \Rightarrow \phi(x_i) \ge \phi(c) + m(x_i-c).$$
This proves that indeed all $(x_i,\phi(x_i))$ lie at or above the line $y=m(x-c)+\phi(c)$, and the proof now concludes easily:
$$\begin{eqnarray}
\sum_{i=1}^k\lambda_i\phi(x_i) & \ge & \sum_{i=1}^k\lambda_i(\phi(c)+m(x_i-c))\\
& = & \sum_{i=1}^k\lambda_i\phi(c) + m \sum_{i=1}^k\lambda_ix_i - m \sum_{i=1}^k\lambda_ic\\
& = & \phi(c) \sum_{i=1}^k\lambda_i +mc - mc\times1\\
& = & \phi(c)\\
& = & \phi(\sum_{i=1}^k\lambda_ix_i),\\
\end{eqnarray}$$
which is ecactly the statement to prove.
Best Answer
The function $f$ has full domain and is thus automatically continuous (see Rockafeallar's Convex Analysis). So the result follows from the Intermediate Value Theorem (connect on a line segment $x_i$ and $x_j$, where these two take on the minimum and the maximum).